o) Ta có
$\sin x + \cos(3x) = \cos(2x)$
$<-> \cos(3x) - \cos(2x) = -\sin(x)$
Đặt $x = 2t$. Khi đó ptrinh trở thành
$\cos(6t) - \cos(4t) = -\sin(2t)$
$<-> -2\sin(5t) \sin(t) = -2\sin t \cos t$
$<-> \sin t[\sin(5t) - \cos t] = 0$
Vậy $\sin t = 0$ hoặc $\sin(5t) = \cos t$
TH1: $\sin t = 0$
Suy ra $t = k\pi$ hay $x = 2k\pi$.
TH2: $\sin(5t) = \cos t$
Tương đương vs
$\cos \left( \dfrac{\pi}{2} - 5t \right) = \cos t$
$<-> \dfrac{\pi}{2} - 5t = \pm t + 2k\pi$
Vậy $2t = x = \dfrac{\pi}{6} + \dfrac{2k\pi}{3}$ hoặc $2t = x = \dfrac{\pi}{4} + k\pi$
Vậy $S = \left\{ 2k\pi, \dfrac{\pi}{6} + \dfrac{2k\pi}{3}, \dfrac{\pi}{4} + k\pi \right\}$.
