Additional Mathematics 0606 / Trigonometry / 1.331.33: Cosine–tangent to quadratic in sine0606/13/O/N/20 — Question 11 · 7 marksMark as done · Save for later · Show all solutions(a)Given that 2cosx=3tanx2\cos x = 3\tan x2cosx=3tanx, show that 2sin2x+3sinx−2=02\sin^{2}x + 3\sin x - 2 = 02sin2x+3sinx−2=0.[3]▸ Answer(b)Hence solve 2cos(2α+π4)=3tan(2α+π4)2\cos\left(2\alpha + \tfrac{\pi}{4}\right) = 3\tan\left(2\alpha + \tfrac{\pi}{4}\right)2cos(2α+4π)=3tan(2α+4π) for 0<α<π0 < \alpha < \pi0<α<π radians, giving your answers in terms of π\piπ.[4]▸ Answer▸ Official mark scheme← 1.21: Cosine graph: amplitude, period and sketch2.12: Exact tan 15° and Pythagoras →
