en 'Poly Root Finder'. Set Degree to 3. Enter $a_3 = 4$, $a_2 = 0$, $a_1 = -3$, $a_0 = 0.85$. Press [SOLVE]."},"desmos":{"expression":"y = 4x^3 - 3x + 0.85","instructions":"Type $y = 4x^3 - 3x + 0.85$ and find the $x$-intercepts."},"description":"Solve the cubic polynomial $4x^3 - 3x + 0.85 = 0$."}]},{"type":"text","order":2,"title":"Find the angles for each root","content":"Now we find the angles $A$ in the interval $0 \\leq A \\leq 180^{\\circ}$ for each root.\n\nSince sine is positive in the first two quadrants, any negative root $x < 0$ will yield no valid solutions within our interval $[0^{\\circ}, 180^{\\circ}]$. Let's calculate the angles for the positive roots:\n\n**Case 1: $x_1 \\approx 0.33244$**\n\n$$A_1 = \\sin^{-1}(0.33244) \\approx 19.41^{\\circ}$$\n\n$$A_2 = 180^{\\circ} - 19.41^{\\circ} \\approx 160.59^{\\circ}$$\n\n**Case 2: $x_2 \\approx 0.65079$**\n\n$$A_3 = \\sin^{-1}(0.65079) \\approx 40.60^{\\circ}$$\n\n$$A_4 = 180^{\\circ} - 40.60^{\\circ} \\approx 139.40^{\\circ}$$\n\n**Case 3: $x_3 \\approx -0.98323$**\n\nSince this value is negative, there are no solutions for $A$ in the interval $[0^{\\circ}, 180^{\\circ}]$."},{"type":"text","order":3,"title":"State the final solutions","content":"Rounding each of our valid angles to 1 decimal place as requested, we get:\n\n$$A = 19.4^{\\circ}, \\; 40.6^{\\circ}, \\; 139.4^{\\circ}, \\; 160.6^{\\circ}$$","highlights":[{"text":"Give your answers to 1 decimal place.","location":"specification"}]}]}],"generatedAt":"2026-06-09T03:11:04.872Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":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,"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"},"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":"Facto

Trigonometry and Modelling

Question 8

Trigonometry and Modelling Questions

Trigonometry and Modelling

Question 8

Trigonometry and Modelling Questions