Target Mathematics

1.71: Trigonometric identities and equations

4PM1/2/January/2017 — Question 2 · 7 marks

(a) Show that the equation 6cos⁡2α−sin⁡α=56\cos^2\alpha-\sin\alpha=5 can be written as
6sin⁡2α+sin⁡α−1=06\sin^2\alpha+\sin\alpha-1=0
(2)
(b) Solve, to 1 decimal place where appropriate, for 0≤θ≤900\le\theta\le90,
6cos⁡2(2θ+40)∘−sin⁡(2θ+40)∘=56\cos^2(2\theta+40)^\circ-\sin(2\theta+40)^\circ=5
(5)