Additional Mathematics 0606 / Simultaneous equations and quadratics / 3.13.1: Sine equation with phase shift in radians0606/11/M/J/24 — Question 8 · 5 marksMark as done · Save for later · Show all solutionsSolve the equation 4sin2 (2α−π3)=14\sin^{2}\!\left(2\alpha - \dfrac{\pi}{3}\right) = 14sin2(2α−3π)=1 for −π2⩽α⩽π2-\dfrac{\pi}{2} \leqslant \alpha \leqslant \dfrac{\pi}{2}−2π⩽α⩽2π. Give your answers in terms of π\piπ.[5]▸ Answer▸ Official mark scheme← 3.18: Power simultaneous equations in xxx and yyy3.2: Exponential simultaneous equations and reciprocal exponential →
