Skip to content
MathsGenie logo
Open app

Course home

  1. AS Level
  3. Maths WJEC
  5. Question bank

Trigonometric Identities and Equations

MediumHard
12345678910111213141516171819202122232425
Question 19

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

  1. AS Level
  2. /Maths
  3. /Trigonometric Identities and Equations