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$\\sin x = 3$) M1\n- Final answers: $x = -30^\\circ, -150^\\circ$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Attempt to substitute identity $\\cos 2x = 1 - 2 \\sin^2 x$.","code":"M"},{"id":"p2","marks":1,"description":"Formation of a correct 3-term quadratic equation in $\\sin x$.","code":"M"},{"id":"p3","marks":1,"description":"Correct factorization or solution method for their quadratic equation.","code":"M"},{"id":"p4","marks":1,"description":"Identification of $\\sin x = -1/2$ and explicit or implied rejection of the other root.","code":"M"},{"id":"p5","marks":1,"description":"Final answers $x = -30^\\circ$ and $x = -150^\\circ$ provided correctly in range.","code":"A"}],"finalAnswers":["-30^\\circ, -150^\\circ"],"generatedAt":"2026-05-06T14:04:18.884Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic_quadratic","label":"Algebraic Substitution and Quadratic Formula","steps":[{"type":"text","order":0,"title":"Substitute double-angle identity","content":"To solve the equation:\n\n$$\\cos 2x + 5 \\sin x + 2 = 0$$\n\nwe first need to write it in terms of a single trigonometric function. Since there is a $\\sin x$ term, we use the double-angle identity for $\\cos 2x$ that involves sine:\n\n$$\\cos 2x = 1 - 2 \\sin^2 x$$\n\nSubstituting this into our equation gives:\n\n$$(1 - 2\\sin^2 x) + 5\\sin x + 2 = 0$$\n\nThis earns the first method mark **M1** for correct substitution."},{"type":"text","order":1,"title":"Form a quadratic equation","content":"Now, we simplify and rearrange the equation to form a standard quadratic equation in terms of $\\sin x$.\n\nCombining the constant terms $1$ and $2$:\n\n$$-2 \\sin^2 x + 5 \\sin x + 3 = 0$$\n\nTo make it easier to factorise, multiply the entire equation by $-1$:\n\n$$2 \\sin^2 x - 5 \\sin x - 3 = 0$$\n\nThis earns the second method mark **M1** for forming the quadratic."},{"type":"text","order":2,"title":"Factorise the quadratic","content":"To solve $2 \\sin^2 x - 5 \\sin x - 3 = 0$, we can think of this as a quadratic of the form $2y^2 - 5y - 3 = 0$, where $y = \\sin x$.\n\nWe look for two numbers that multiply to $2 \\times (-3) = -6$ and add to $-5$. These numbers are $-6$ and $1$. Splitting the middle term:\n\n$$2y^2 - 6y + y - 3 = 0$$\n\n$$2y(y - 3) + 1(y - 3) = 0$$\n\n$$(2y + 1)(y - 3) = 0$$\n\nSubstituting $y = \\sin x$ back in gives:\n\n$$(2 \\sin x + 1)(\\sin x - 3) = 0$$\n\nThis earns the third method mark **M1** for correct factorisation."},{"type":"text","order":3,"title":"Solve for sin x","content":"Now set each factor to zero to find the possible values for $\\sin x$:\n\n1. From $2 \\sin x + 1 = 0$:\n\n$$\\sin x = -1/2$$\n\n2. From $\\sin x - 3 = 0$:\n\n$$\\sin x = 3$$\n\nSince the range of the sine function is $-1 \\leq \\sin x \\leq 1$, there are no real solutions to $\\sin x = 3$. Therefore, we reject this case.\n\nThis earns the fourth method mark **M1** for identifying $\\sin x = -1/2$ and rejecting $\\sin x = 3$."},{"type":"text","order":4,"title":"Find the final angles","content":"Now we solve $\\sin x = -1/2$ within the interval $-180^\\circ \\leq x \\leq 180^\\circ$.\n\nFirst, find the principal value (often called the reference angle) using your calculator:\n\n$$\\sin^{-1}(1/2) = 30^\\circ$$\n\nSince $\\sin x$ is negative ($-1/2$), the solutions must lie in the third and fourth quadrants.\n\nUsing the symmetry of the sine graph within our interval:\n- The solution in the fourth quadrant is:\n $$x = -30^\\circ$$\n- The solution in the third quadrant is:\n $$x = -180^\\circ - (-30^\\circ) = -150^\\circ$$\n\nBoth values lie within the required interval $-180^\\circ \\leq x \\leq 180^\\circ$.\n\nTherefore, our final answers are:\n\n$$x = -30^\\circ, -150^\\circ$$\n\nThis earns the final accuracy mark **A1**."}]},{"id":"graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Define the function","content":"To solve the equation $\\cos 2x + 5 \\sin x + 2 = 0$ graphically, we define the left-hand side as a function of $x$:\n\n$$y = \\cos 2x + 5 \\sin x + 2$$\n\nSolving the equation is equivalent to finding the x-intercepts (where $y = 0$) of this function in the interval $[-180^\\circ, 180^\\circ]$. Since the interval is given in degrees, make sure your graphing tool or calculator is set to **Degree** mode."},{"type":"text","order":1,"title":"Set calculator window","content":"To get a clear view of the function within the specified domain, configure your graphing window with the following settings:\n\n- **x-Min:** $-180$\n- **x-Max:** $180$\n- **x-Scale:** $30$ (to easily identify degree intervals)\n- **y-Min:** $-5$\n- **y-Max:** $10$\n\nNow, input the equation `y = cos(2x) + 5sin(x) + 2` into your calculator."},{"type":"text","order":2,"title":"Locate the roots","content":"Observe the graph on your screen. You will see that the curve intersects the x-axis at two points within the interval $[-180^\\circ, 180^\\circ]$.\n\nTo find these precise values:\n- Use the **Zero** or **Intercept** tool on your calculator to calculate the coordinates of these intersection points.\n- By selecting a left bound and right bound around each intersection, the calculator computes the exact roots:\n - First root: $x = -150^\\circ$\n - Second root: $x = -30^\\circ$"},{"type":"text","order":3,"title":"State final solutions","content":"Both of these values lie well within the specified domain of $[-180^\\circ, 180^\\circ]$.\n\nTherefore, the solutions to the equation are:\n\n$$x = -150^\\circ, \\quad x = -30^\\circ$$"}]}],"generatedAt":"2026-06-09T02:29:13.855Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"fe4b2134-1820-4cde-bdab-163ab7e3da0d:algebraic_quadratic","label":"Algebraic Substitution and Quadratic Formula","steps":[{"type":"text","order":0,"title":"Substitute double-angle identity","content":"To solve the equation:\n\n$$\\cos 2x + 5 \\sin x + 2 = 0$$\n\nwe first need to write it in terms of a single trigonometric function. Since there is a $\\sin x$ term, we use the double-angle identity for $\\cos 2x$ that involves sine:\n\n$$\\cos 2x = 1 - 2 \\sin^2 x$$\n\nSubstituting this into our equation gives:\n\n$$(1 - 2\\sin^2 x) + 5\\sin x + 2 = 0$$\n\nThis earns the first method mark **M1** for correct substitution."},{"type":"text","order":1,"title":"Form a quadratic equation","content":"Now, we simplify and rearrange the equation to form a standard quadratic equation in terms of $\\sin x$.\n\nCombining the constant terms $1$ and $2$:\n\n$$-2 \\sin^2 x + 5 \\sin x + 3 = 0$$\n\nTo make it easier to factorise, multiply the entire equation by $-1$:\n\n$$2 \\sin^2 x - 5 \\sin x - 3 = 0$$\n\nThis earns the second method mark **M1** for forming the quadratic."},{"type":"text","order":2,"title":"Factorise the quadratic","content":"To solve $2 \\sin^2 x - 5 \\sin x - 3 = 0$, we can think of this as a quadratic of the form $2y^2 - 5y - 3 = 0$, where $y = \\sin x$.\n\nWe look for two numbers that multiply to $2 \\times (-3) = -6$ and add to $-5$. These numbers are $-6$ and $1$. Splitting the middle term:\n\n$$2y^2 - 6y + y - 3 = 0$$\n\n$$2y(y - 3) + 1(y - 3) = 0$$\n\n$$(2y + 1)(y - 3) = 0$$\n\nSubstituting $y = \\sin x$ back in gives:\n\n$$(2 \\sin x + 1)(\\sin x - 3) = 0$$\n\nThis earns the third method mark **M1** for correct factorisation."},{"type":"text","order":3,"title":"Solve for sin x","content":"Now set each factor to zero to find the possible values for $\\sin x$:\n\n1. From $2 \\sin x + 1 = 0$:\n\n$$\\sin x = -1/2$$\n\n2. From $\\sin x - 3 = 0$:\n\n$$\\sin x = 3$$\n\nSince the range of the sine function is $-1 \\leq \\sin x \\leq 1$, there are no real solutions to $\\sin x = 3$. Therefore, we reject this case.\n\nThis earns the fourth method mark **M1** for identifying $\\sin x = -1/2$ and rejecting $\\sin x = 3$."},{"type":"text","order":4,"title":"Find the final angles","content":"Now we solve $\\sin x = -1/2$ within the interval $-180^\\circ \\leq x \\leq 180^\\circ$.\n\nFirst, find the principal value (often called the reference angle) using your calculator:\n\n$$\\sin^{-1}(1/2) = 30^\\circ$$\n\nSince $\\sin x$ is negative ($-1/2$), the solutions must lie in the third and fourth quadrants.\n\nUsing the symmetry of the sine graph within our interval:\n- The solution in the fourth quadrant is:\n $$x = -30^\\circ$$\n- The solution in the third quadrant is:\n $$x = -180^\\circ - (-30^\\circ) = -150^\\circ$$\n\nBoth values lie within the required interval $-180^\\circ \\leq x \\leq 180^\\circ$.\n\nTherefore, our final answers are:\n\n$$x = -30^\\circ, -150^\\circ$$\n\nThis earns the final accuracy mark **A1**."}]},{"id":"fe4b2134-1820-4cde-bdab-163ab7e3da0d:graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Define the function","content":"To solve the equation $\\cos 2x + 5 \\sin x + 2 = 0$ graphically, we define the left-hand side as a function of $x$:\n\n$$y = \\cos 2x + 5 \\sin x + 2$$\n\nSolving the equation is equivalent to finding the x-intercepts (where $y = 0$) of this function in the interval $[-180^\\circ, 180^\\circ]$. Since the interval is given in degrees, make sure your graphing tool or calculator is set to **Degree** mode."},{"type":"text","order":1,"title":"Set calculator window","content":"To get a clear view of the function within the specified domain, configure your graphing window with the following settings:\n\n- **x-Min:** $-180$\n- **x-Max:** $180$\n- **x-Scale:** $30$ (to easily identify degree intervals)\n- **y-Min:** $-5$\n- **y-Max:** $10$\n\nNow, input the equation `y = cos(2x) + 5sin(x) + 2` into your calculator."},{"type":"text","order":2,"title":"Locate the roots","content":"Observe the graph on your screen. You will see that the curve intersects the x-axis at two points within the interval $[-180^\\circ, 180^\\circ]$.\n\nTo find these precise values:\n- Use the **Zero** or **Intercept** tool on your calculator to calculate the coordinates of these intersection points.\n- By selecting a left bound and right bound around each intersection, the calculator computes the exact roots:\n - First root: $x = -150^\\circ$\n - Second root: $x = -30^\\circ$"},{"type":"text","order":3,"title":"State final solutions","content":"Both of these values lie well within the specified domain of $[-180^\\circ, 180^\\circ]$.\n\nTherefore, the solutions to the equation are:\n\n$$x = -150^\\circ, \\quad x = -30^\\circ$$"}]}],"generatedAt":"2026-06-09T02:29:50.859Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":27},"walkthroughVideo":null},{"id":"4ed7094e-cf03-4e1e-8691-1d3d0175a970","revisionId":"de27527b-7861-4ecc-b905-480729bd71eb","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"2dd0e67c-5b44-42a8-b80f-3e556de903a9","label":"a","content":"Solve, for $-180 \\leq x \\leq 180$, the equation,\n\n$$\\cos 2x + 3 \\cos x - 1 = 0$$","markscheme":"- Substitutes $\\cos 2x = 2 \\cos^2 x - 1$ M1\n- Forms quadratic: $2 \\cos^2 x + 3 \\cos x - 2 = 0$ M1\n- Factorises to $(2 \\cos x - 1)(\\cos x + 2) = 0$ M1\n- Solves $\\cos x = \\frac{1}{2}$ (rejects $\\cos x = -2$) M1\n- Final answers: $x = 60^\\circ, -60^\\circ$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Attempt to substitute identity $\\cos 2x = 2 \\cos^2 x - 1$.","code":"M"},{"id":"p2","marks":1,"description":"Formation of a correct 3-term quadratic equation in $\\cos x$.","code":"M"},{"id":"p3","marks":1,"description":"Correct factorization or solution method for their quadratic equation.","code":"M"},{"id":"p4","marks":1,"description":"Identification of $\\cos x = 1/2$ and explicit or implied rejection of the other root.","code":"M"},{"id":"p5","marks":1,"description":"Final answers $x = 60^\\circ$ and $x = -60^\\circ$ provided correctly in range.","code":"A"}],"finalAnswers":["60^\\circ, -60^\\circ"],"generatedAt":"2026-05-06T14:04:20.788Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Use double-angle identity","content":"To solve the equation $\\cos 2x + 3 \\cos x - 1 = 0$, we first need to express all trigonometric terms in terms of a single angle, $x$.\n\nWe can use the double-angle identity for cosine from our formula sheet:\n$$\\cos 2x = 2 \\cos^2 x - 1$$\n\nSubstituting this into our equation gives:\n$$(2 \\cos^2 x - 1) + 3 \\cos x - 1 = 0$$"},{"type":"text","order":1,"title":"Form a quadratic equation","content":"Next, we simplify the terms to form a standard quadratic equation in terms of $\\cos x$:\n\n$$2 \\cos^2 x + 3 \\cos x - 2 = 0$$\n\nTo make this easier to solve, we can use algebraic substitution. Let $y = \\cos x$. This transforms our equation into a standard quadratic equation:\n\n$$2y^2 + 3y - 2 = 0$$"},{"type":"text","order":2,"title":"Factorise the quadratic","content":"Now we factorise the quadratic equation $2y^2 + 3y - 2 = 0$.\n\nWe look for two numbers that multiply to $2 \\times (-2) = -4$ and add to $3$. These numbers are $4$ and $-1$.\n\nRewriting and factorising by grouping:\n$$\\begin{aligned} 2y^2 + 4y - y - 2 &= 0 \\\\ 2y(y + 2) - 1(y + 2) &= 0 \\\\ (2y - 1)(y + 2) &= 0 \\end{aligned}$$\n\nSubstituting $\\cos x$ back in for $y$, we get:\n$$(2 \\cos x - 1)(\\cos x + 2) = 0$$"},{"type":"text","order":3,"title":"Solve for cos x","content":"Setting each factor to zero gives two possible branches:\n\n1. $$2 \\cos x - 1 = 0 \\implies \\cos x = \\frac{1}{2}$$\n2. $$\\cos x + 2 = 0 \\implies \\cos x = -2$$\n\nSince the range of the cosine function is $-1 \\leq \\cos x \\leq 1$, the equation $\\cos x = -2$ has no real solutions. Thus, we reject this branch and focus solely on:\n$$\\cos x = \\frac{1}{2}$$"},{"type":"text","order":4,"title":"Find the final angles","content":"We now find the values of $x$ in the interval $-180^\\circ \\leq x \\leq 180^\\circ$ that satisfy $\\cos x = \\frac{1}{2}$.\n\nThe principal value is:\n$$x = \\arccos\\left(\\frac{1}{2}\\right) = 60^\\circ$$\n\nUsing the symmetry of the cosine graph, $\\cos(-x) = \\cos(x)$, the second solution within our domain is:\n$$x = -60^\\circ$$\n\nBoth values lie within the required interval $[-180^\\circ, 180^\\circ]$.\n\nTherefore, the final solutions are:\n$$x = 60^\\circ, -60^\\circ$$"}]},{"id":"graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Define the function","content":"To solve the equation using a graphical approach, we define a function $y = f(x)$ representing the left side of the equation:\n\n$$y = \\cos(2x) + 3\\cos(x) - 1$$\n\nWe need to find the $x$-intercepts (roots) of this function where $y = 0$, within the specified interval:\n\n$$-180^\\circ ≤ x ≤ 180^\\circ$$"},{"type":"text","order":1,"title":"Set up your calculator","content":"Since the interval is specified in degrees (from $-180^\\circ$ to $180^\\circ$), make sure your graphing calculator is set to **Degree mode** before entering the function and adjusting the window settings.","calculator":[{"gdc":{"expression":"cos(2x) + 3cos(x) - 1","instructions":"1. Go to Graph mode and set Angle to 'Degree'.\n2. Enter the function: y1 = cos(2x) + 3cos(x) - 1.\n3. Set the View-Window (V-Window) for x to [-180, 180] and y to [-5, 5].\n4. Draw the graph."},"ti84":{"expression":"cos(2X) + 3cos(X) - 1","instructions":"1. Press [MODE] and ensure 'DEGREE' is highlighted.\n2. Press [Y=] and enter Y1 = cos(2X) + 3cos(X) - 1.\n3. Press [WINDOW] and set:\n - Xmin = -180\n - Xmax = 180\n - Xscl = 30\n - Ymin = -5\n - Ymax = 5\n4. Press [GRAPH]."},"desmos":{"expression":"y = \\cos(2x) + 3\\cos(x) - 1","instructions":"1. Click on the Graph Settings wrench icon in the top right corner.\n2. Select 'Degrees'.\n3. Type in the equation: y = cos(2x) + 3cos(x) - 1.\n4. Set the X-axis bounds to [-180, 180] and the Y-axis bounds to [-5, 5]."},"description":"Configure your calculator to graph the function $y = \\cos(2x) + 3\\cos(x) - 1$ in degree mode."}]},{"type":"text","order":2,"title":"Locate the x-intercepts","content":"Looking at the generated graph, you will see a symmetric curve with two $x$-intercepts within the interval $[-180^\\circ, 180^\\circ]$.\n\nWe can find these precise points using the zero/root-finding feature of the graphing tool:\n\n- The first intercept on the negative side is located at:\n $$x = -60^\\circ$$\n\n- The second intercept on the positive side is located at:\n $$x = 60^\\circ$$","calculator":[{"gdc":{"instructions":"1. Press [G-Solve] (usually [F5]).\n2. Select [Root] (usually [F1]).\n3. Use the arrow keys to toggle between the two solutions: x = -60° and x = 60°."},"ti84":{"instructions":"1. Press [2nd] then [CALC] (above [TRACE]) and select '2: zero'.\n2. For the left root: Set a Left Bound to the left of -60 (e.g., -90), a Right Bound to the right (e.g., -30), and press [ENTER] to guess. The calculator displays X = -60.\n3. Repeat for the right root: Set a Left Bound (e.g., 30), a Right Bound (e.g., 90), and press [ENTER] to guess. The calculator displays X = 60."},"desmos":{"instructions":"1. Simply click on the points where the curve crosses the x-axis.\n2. Desmos will highlight the coordinates as grey dots, displaying (-60, 0) and (60, 0)."},"description":"Find the roots of the equation on your graphing tool."}]},{"type":"text","order":3,"title":"State the final solutions","content":"The values of $x$ in the interval $[-180^\\circ, 180^\\circ]$ that satisfy the equation are:\n\n$$x = -60^\\circ \\quad \\text{and} \\quad x = 60^\\circ$$"}]}],"generatedAt":"2026-06-09T05:35:26.170Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"2dd0e67c-5b44-42a8-b80f-3e556de903a9:algebraic","label":"Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Use double-angle identity","content":"To solve the equation $\\cos 2x + 3 \\cos x - 1 = 0$, we first need to express all trigonometric terms in terms of a single angle, $x$.\n\nWe can use the double-angle identity for cosine from our formula sheet:\n$$\\cos 2x = 2 \\cos^2 x - 1$$\n\nSubstituting this into our equation gives:\n$$(2 \\cos^2 x - 1) + 3 \\cos x - 1 = 0$$"},{"type":"text","order":1,"title":"Form a quadratic equation","content":"Next, we simplify the terms to form a standard quadratic equation in terms of $\\cos x$:\n\n$$2 \\cos^2 x + 3 \\cos x - 2 = 0$$\n\nTo make this easier to solve, we can use algebraic substitution. Let $y = \\cos x$. This transforms our equation into a standard quadratic equation:\n\n$$2y^2 + 3y - 2 = 0$$"},{"type":"text","order":2,"title":"Factorise the quadratic","content":"Now we factorise the quadratic equation $2y^2 + 3y - 2 = 0$.\n\nWe look for two numbers that multiply to $2 \\times (-2) = -4$ and add to $3$. These numbers are $4$ and $-1$.\n\nRewriting and factorising by grouping:\n$$\\begin{aligned} 2y^2 + 4y - y - 2 &= 0 \\\\ 2y(y + 2) - 1(y + 2) &= 0 \\\\ (2y - 1)(y + 2) &= 0 \\end{aligned}$$\n\nSubstituting $\\cos x$ back in for $y$, we get:\n$$(2 \\cos x - 1)(\\cos x + 2) = 0$$"},{"type":"text","order":3,"title":"Solve for cos x","content":"Setting each factor to zero gives two possible branches:\n\n1. $$2 \\cos x - 1 = 0 \\implies \\cos x = \\frac{1}{2}$$\n2. $$\\cos x + 2 = 0 \\implies \\cos x = -2$$\n\nSince the range of the cosine function is $-1 \\leq \\cos x \\leq 1$, the equation $\\cos x = -2$ has no real solutions. Thus, we reject this branch and focus solely on:\n$$\\cos x = \\frac{1}{2}$$"},{"type":"text","order":4,"title":"Find the final angles","content":"We now find the values of $x$ in the interval $-180^\\circ \\leq x \\leq 180^\\circ$ that satisfy $\\cos x = \\frac{1}{2}$.\n\nThe principal value is:\n$$x = \\arccos\\left(\\frac{1}{2}\\right) = 60^\\circ$$\n\nUsing the symmetry of the cosine graph, $\\cos(-x) = \\cos(x)$, the second solution within our domain is:\n$$x = -60^\\circ$$\n\nBoth values lie within the required interval $[-180^\\circ, 180^\\circ]$.\n\nTherefore, the final solutions are:\n$$x = 60^\\circ, -60^\\circ$$"}]},{"id":"2dd0e67c-5b44-42a8-b80f-3e556de903a9:graphical","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Define the function","content":"To solve the equation using a graphical approach, we define a function $y = f(x)$ representing the left side of the equation:\n\n$$y = \\cos(2x) + 3\\cos(x) - 1$$\n\nWe need to find the $x$-intercepts (roots) of this function where $y = 0$, within the specified interval:\n\n$$-180^\\circ ≤ x ≤ 180^\\circ$$"},{"type":"text","order":1,"title":"Set up your calculator","content":"Since the interval is specified in degrees (from $-180^\\circ$ to $180^\\circ$), make sure your graphing calculator is set to **Degree mode** before entering the function and adjusting the window settings.","calculator":[{"gdc":{"expression":"cos(2x) + 3cos(x) - 1","instructions":"1. Go to Graph mode and set Angle to 'Degree'.\n2. Enter the function: y1 = cos(2x) + 3cos(x) - 1.\n3. Set the View-Window (V-Window) for x to [-180, 180] and y to [-5, 5].\n4. Draw the graph."},"ti84":{"expression":"cos(2X) + 3cos(X) - 1","instructions":"1. Press [MODE] and ensure 'DEGREE' is highlighted.\n2. Press [Y=] and enter Y1 = cos(2X) + 3cos(X) - 1.\n3. Press [WINDOW] and set:\n - Xmin = -180\n - Xmax = 180\n - Xscl = 30\n - Ymin = -5\n - Ymax = 5\n4. Press [GRAPH]."},"desmos":{"expression":"y = \\cos(2x) + 3\\cos(x) - 1","instructions":"1. Click on the Graph Settings wrench icon in the top right corner.\n2. Select 'Degrees'.\n3. Type in the equation: y = cos(2x) + 3cos(x) - 1.\n4. Set the X-axis bounds to [-180, 180] and the Y-axis bounds to [-5, 5]."},"description":"Configure your calculator to graph the function $y = \\cos(2x) + 3\\cos(x) - 1$ in degree mode."}]},{"type":"text","order":2,"title":"Locate the x-intercepts","content":"Looking at the generated graph, you will see a symmetric curve with two $x$-intercepts within the interval $[-180^\\circ, 180^\\circ]$.\n\nWe can find these precise points using the zero/root-finding feature of the graphing tool:\n\n- The first intercept on the negative side is located at:\n $$x = -60^\\circ$$\n\n- The second intercept on the positive side is located at:\n $$x = 60^\\circ$$","calculator":[{"gdc":{"instructions":"1. Press [G-Solve] (usually [F5]).\n2. Select [Root] (usually [F1]).\n3. Use the arrow keys to toggle between the two solutions: x = -60° and x = 60°."},"ti84":{"instructions":"1. Press [2nd] then [CALC] (above [TRACE]) and select '2: zero'.\n2. For the left root: Set a Left Bound to the left of -60 (e.g., -90), a Right Bound to the right (e.g., -30), and press [ENTER] to guess. The calculator displays X = -60.\n3. Repeat for the right root: Set a Left Bound (e.g., 30), a Right Bound (e.g., 90), and press [ENTER] to guess. The calculator displays X = 60."},"desmos":{"instructions":"1. Simply click on the points where the curve crosses the x-axis.\n2. Desmos will highlight the coordinates as grey dots, displaying (-60, 0) and (60, 0)."},"description":"Find the roots of the equation on your graphing tool."}]},{"type":"text","order":3,"title":"State the final solutions","content":"The values of $x$ in the interval $[-180^\\circ, 180^\\circ]$ that satisfy the equation are:\n\n$$x = -60^\\circ \\quad \\text{and} \\quad x = 60^\\circ$$"}]}],"generatedAt":"2026-06-09T05:35:34.356Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"0aec2d07-eeb2-4854-99e5-78c5846bc99a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"4ed7094e-cf03-4e1e-8691-1d3d0175a970","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"2cb328b9-19c8-4983-a09d-31ba9ff0ece8","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"02abd9f1-3793-4eae-9173-31c3de431a06","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"16625c96-dbb8-4a12-aaca-d77403e7dbac","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"1eeb206a-a348-4e3a-a08f-bbe35d6be702","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"245935ae-b393-4618-b200-7ceb22ea40be","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"2867863b-b9e7-4f84-ae17-8331b1c6c9e5","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"2a2d5a62-4963-473f-a4bb-a7e8dd56da56","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"2dd44c3f-bdf0-42d9-bd30-910ce4b32418","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"38c35132-2384-4b6c-8845-8f63f74ad764","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"3ac9e0d1-6515-488d-b25f-aca54b9664e0","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"3cc96763-3087-4987-be28-0948c2f5f671","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"513e051a-a107-415f-a26f-5a645b94caae","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"515db5da-764b-47d7-8bd9-8ce0ec7e690f","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"5326fecf-c197-46ff-a3b1-eb44cf54fd0e","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"5dbd6a16-7dfb-44ce-aece-fe9cee1738bd","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"6241cfd3-b352-4d84-ba16-859861bec3b8","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"70499ec1-c56f-420f-adde-ce7a7491e875","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"85a5a693-25a7-4a40-8b84-3e96aa4a8a7b","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"8a1fa494-fdf6-48ec-a9d8-b6866980397d","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"8d4206fd-76f3-46f8-98e1-08940aaef0f4","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"94a4665e-3a10-44ef-9810-326acd4c49d0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"99044bf8-6ee5-436d-aa4a-345d00112f02","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"9e134eed-4e06-4005-804b-e37007bf9af1","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"9ea04f6f-ce65-4283-8cff-0c256fc188da","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"b658e46a-2d28-4b3d-8c6d-677b07f5d6db","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"d5905b02-7f0f-48a0-9d61-39baaede11c7","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"e9fd3e58-1fc3-415c-8bfb-44bd919805a6","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true}],"title":"7.3 Solving Trigonometric Equations","topicId":"4fd79770-a76b-425c-861f-a7adb80ec9ca","topicSidebar":"$L3a","topicSlug":"a-level-pure-mathematics-7-3-solving-trigonometric-equations","topicTitle":"7.3 Solving Trigonometric Equations"}]}],"$L3b","$L3c"]}]