Trigonometric Identities and Equations

MediumHard
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Question 6
Medium65%
a.

Show that the equation

2sin⁡2x=7cos⁡x+5 2\sin^2 x = 7\cos x + 5 2sin2x=7cosx+5

Can be written in the form

2cos⁡2x+7cos⁡x+3=0 2\cos^2 x + 7\cos x + 3 = 0 2cos2x+7cosx+3=0
[3]
b.

Hence solve, for 0≤x<360∘0 \leq x < 360^\circ0≤x<360∘, the equation,

2sin⁡2x=7cos⁡x+5 2\sin^2 x = 7\cos x + 5 2sin2x=7cosx+5
[5]

Trigonometric Identities and Equations Questions

  1. AS Level
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