Show that $\sin{2x} + \cos{2x} - 1 = 2\sin{x} (\cos{x} - \sin{x})$. [2] Hence or otherwise, solve

Question

Show that $\sin{2x} + \cos{2x} - 1 = 2\sin{x} (\cos{x} - \sin{x})$. [2] Hence or otherwise, solve $\sin{2x} + \cos{2x} - 1 + \cos{x} - \sin{x} = 0$ for $0 < x < 2\pi$. [6]

Accepted Answer

(b) $x = \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{\pi}{4}, \frac{5\pi}{4}$

2021 May TZ1 P1 Q6

2021 May TZ1 P1 Q6