Sinx + sin2x= cosx + cos2x

Đáp án:

Giải thích các bước giải: \[\begin{array}{l} \sin x + \sin 2x = \cos x + \cos 2x\\ \Leftrightarrow \sin x - \cos 2x = \cos x - 2\sin x\\ \Leftrightarrow \sin x + 2{\sin ^2}x - 1 = \cos x - 2\sin x.\cos x\\ \Leftrightarrow (2\sin x - 1)(\sin x + 1) - \cos x(1 - 2\sin x) = 0\\ \Leftrightarrow (2\sin x - 1)(\sin x + \cos x + 1) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{6} + k2\pi \\ x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.\\ \sin x + \cos x = - 1 \Leftrightarrow \sin (x + \frac{\pi }{4}) = \frac{{ - \sqrt 2 }}{2} \Leftrightarrow \left[ \begin{array}{l} x = \frac{{ - \pi }}{2} + k2\pi \\ x = \pi + k2\pi \end{array} \right. \end{array} \right. \end{array}\]