题目 Problem (a) Prove the identity sin2x−cosx−11+cosx=−cosx\frac{\sin^2 x - \cos x - 1}{1 + \cos x} = -\cos x1+cosxsin2x−cosx−1=−cosx [3] (b) Hence solve the equation sin2x−cosx−12+2cosx=14\frac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \frac142+2cosxsin2x−cosx−1=41 for 0∘≤x≤360∘0^\circ \leq x \leq 360^\circ0∘≤x≤360∘. [3] 题目中文翻译 (a) 证明恒等式 sin2x−cosx−11+cosx=−cosx\frac{\sin^2 x - \cos x - 1}{1 + \cos x} = -\cos x1+cosxsin2x−cosx−1=−cosx [3] (b) 由此解方程 sin2x−cosx−12+2cosx=14\frac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \frac142+2cosxsin2x−cosx−1=41 其中 0∘≤x≤360∘0^\circ \leq x \leq 360^\circ0∘≤x≤360∘。 [3] 解答
