Solve the inequality $$\frac{(2x+1)(x-2)}{x-1}\leq0.$$[3] Hence solve, in exact form, \[\frac{(2\cos

Question

Solve the inequality $$\frac{(2x+1)(x-2)}{x-1}\leq0.$$[3] Hence solve, in exact form, $$\frac{(2\cos2\theta+1)(\cos2\theta-2)}{\cos2\theta-1}\leq0,\qquad 0<\theta<\pi.$$[2]

Accepted Answer

(i) $x\in\left(-\infty,-\frac12\right]\cup(1,2]$ (ii) $\frac{\pi}{3}\leq\theta\leq\frac{2\pi}{3}$

Double-angle cosine substitution in a rational inequality

Double-angle cosine substitution in a rational inequality

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