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1.5.15 Trigonometric equations

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Question 28

A student is asked to solve the equation $sin^2 x = 1$ for $\leqslant x \leqslant 360°$ .

The student writes:

Step 1: $sin⁡x=1\sin x = 1$

Step 2: $x = 90°$

Explain the error the student has made in Step 1.

[1]

State the correct solutions of $sin^2 x = 1$ for $\leqslant x \leqslant 360°$ .

[2]

A second student solves the equation $sin⁡xcos⁡x=sin⁡x\sin x\cos x = \sin x$ for $\leqslant x < 360°\,$ by dividing both sides by $sin⁡x\sin x$ , obtaining $cos⁡x=1\cos x = 1$ and hence $x = 0°$ . Explain what is wrong with this method, and give the complete solution set.

[3]

Markscheme

1.5.15 Trigonometric equations Questions

A Level
Maths
1.5.15 Trigonometric equations

42 exam-style questions on OCR A Level Maths 1.5.15 Trigonometric equations. Each one has a worked solution and a mark scheme showing where the marks go.

1.5.15 Trigonometric equations

Question bank

1.5.15 Trigonometric equations

Easy Medium Hard

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

Question 28

A student is asked to solve the equation $sin^2 x = 1$ for $\leqslant x \leqslant 360°$ .

The student writes:

Step 1: $sin⁡x=1\sin x = 1$

Step 2: $x = 90°$

Explain the error the student has made in Step 1.

[1]

State the correct solutions of $sin^2 x = 1$ for $\leqslant x \leqslant 360°$ .

[2]

[3]

Markscheme

1.5.15 Trigonometric equations Questions

A Level
Maths
1.5.15 Trigonometric equations

42 exam-style questions on OCR A Level Maths 1.5.15 Trigonometric equations. Each one has a worked solution and a mark scheme showing where the marks go.