JAMB 1998 · UME · Q34

Solve the equation $\cos x + \sin x = \dfrac{1}{\cos x - \sin x}$ for $0 \le x < 2\pi$ .

Worked solution (try it first)

Multiply both sides by $\cos x - \sin x$ : $(\cos x + \sin x)(\cos x - \sin x) = 1$ , so $\cos^2 x - \sin^2 x = 1$ .
The double-angle formula says $\cos^2 x - \sin^2 x = \cos2x$ , so $\cos2x = 1$ .
For $0 \le x < 2\pi$ , $2x$ runs from 0 to $4\pi$ , and $\cos2x = 1$ at $2x = 0$ or $2\pi$ .
So $x = 0$ or $\pi$ , option D.
(At both, $\cos x - \sin x$ is not zero, so the equation makes sense.)