Skip to content

Course home

1.5.5 Trigonometry

1.5.5 Trigonometry

EasyMediumHard
123456
Question 3
a.

Show that the equation 2sin⁡2x=7cos⁡x+52\sin^2 x = 7\cos x + 52sin2x=7cosx+5 can be written in the form 2cos⁡2x+7cos⁡x+3=02\cos^2 x + 7\cos x + 3 = 02cos2x+7cosx+3=0

[3]
b.

Hence solve, for 0°⩽x<360°0° \leqslant x < 360°0°⩽x<360°, the equation 2sin⁡2x=7cos⁡x+52\sin^2 x = 7\cos x + 52sin2x=7cosx+5.

[5]
Markscheme

1.5.5 Trigonometry Questions

  1. A Level
  2. /Maths
  3. /1.5.5 Trigonometry

42 exam-style questions on WJEC A Level Maths 1.5.5 Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.

Question bank