Additional Mathematics 0606 / Circular measure / 3.403.40: Stationary point and equation for 3sin2x−2cosx3\sin^{2}x - 2\cos x3sin2x−2cosx0606/22/M/J/22 — Question 8 · 10 marksMark as done · Save for later · Show all solutionsThe function fff is defined by f(x)=3sin2x−2cosxf(x) = 3\sin^{2}x - 2\cos xf(x)=3sin2x−2cosx for π2⩽x⩽3π2\dfrac{\pi}{2} \leqslant x \leqslant \dfrac{3\pi}{2}2π⩽x⩽23π, where xxx is in radians.(a)Find the xxx-coordinate of the stationary point on the curve y=f(x)y = f(x)y=f(x).[5]▸ Answer(b)Solve f(x)=1−3cosxf(x) = 1 - 3\cos xf(x)=1−3cosx.[5]▸ Answer▸ Official mark scheme← 3.11: Tangents and sector equal to shaded region3.2: Equal areas in a sector cut by a chord →
