Sin^2 x +sin^2 2x +sin^2 3x =3/2

Đáp án:

$\left[ \begin{array}{l} x = \frac{\pi }{8} + \frac{{k\pi }}{4}\\ x = \frac{\pi }{3} + k\pi \\ x = - \frac{\pi }{3} + k\pi \end{array} \right.$

Giải thích các bước giải: $\begin{array}{l} {\sin ^2}x + {\sin ^2}2x + {\sin ^2}3x = \frac{3}{2}\\ \Leftrightarrow \frac{{1 - \cos 2x}}{2} + \frac{{1 - \cos 4x}}{2} + \frac{{1 - \cos 6x}}{2} = \frac{3}{2}\\ \Leftrightarrow \frac{3}{2} - \frac{1}{2}\left( {\cos 2x + \cos 4x + \cos 6x} \right) = \frac{3}{2}\\ \Leftrightarrow \cos 2x + \cos 4x + \cos 6x = 0\\ \Leftrightarrow \left( {\cos 2x + \cos 6x} \right) + \cos 4x = 0\\ \Leftrightarrow 2\cos 4x\cos 2x + \cos 4x = 0\\ \Leftrightarrow \cos 4x\left( {2\cos 2x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \cos 4x = 0\\ \cos 2x = - \frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 4x = \frac{\pi }{2} + k\pi \\ 2x = \frac{{2\pi }}{3} + k2\pi \\ 2x = - \frac{{2\pi }}{3} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{8} + \frac{{k\pi }}{4}\\ x = \frac{\pi }{3} + k\pi \\ x = - \frac{\pi }{3} + k\pi \end{array} \right. \end{array}$

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