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Trigonometric Identities and Equations

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Question 20

Jacob has to solve the equation

3−sin⁡x=1+2cos⁡2x 3 - \sin x = 1 + 2\cos^2 x 3−sinx=1+2cos2x

where −180∘≤x<180∘-180^{\circ} \leq x < 180^{\circ}−180∘≤x<180∘

Jacob's working is as follows:

3−sin⁡x=1+2cos⁡2x2−sin⁡x=2cos⁡2x2−sin⁡x=2(1−sin⁡2x)2−sin⁡x=2−2sin⁡2x−sin⁡x=−2sin⁡2x1=2sin⁡xsin⁡x=0.5x=30∘ \begin{aligned} 3 - \sin x &= 1 + 2\cos^2 x \\ 2 - \sin x &= 2\cos^2 x \\ 2 - \sin x &= 2(1 - \sin^2 x) \\ 2 - \sin x &= 2 - 2\sin^2 x \\ -\sin x &= -2\sin^2 x \\ 1 &= 2\sin x \\ \sin x &= 0.5 \\ x &= 30^{\circ} \end{aligned} 3−sinx2−sinx2−sinx2−sinx−sinx1sinxx​=1+2cos2x=2cos2x=2(1−sin2x)=2−2sin2x=−2sin2x=2sinx=0.5=30∘​
a.

Explain the two errors that Jacob has made.

[2]
b.

Write down all the values of x x\,x that satisfy the equation

3−sin⁡x=1+2cos⁡2x 3 - \sin x = 1 + 2\cos^2 x 3−sinx=1+2cos2x

where −180∘≤x<180∘-180^{\circ} \leq x < 180^{\circ}−180∘≤x<180∘

[3]

Trigonometric Identities and Equations Questions

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