Đáp án:
\(\left[ \matrix{
x = {{k\pi } \over 2} \hfill \cr
x = {{k\pi } \over 5} \hfill \cr} \right.\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\eqalign{
& \,\,\,\,\,\,{\sin ^2}4x + {\sin ^2}3x = {\sin ^2}2x + {\sin ^2}x \cr
& \Leftrightarrow {{1 - \cos 8x} \over 2} + {{1 - \cos 6x} \over 2} = {{1 - \cos 4x} \over 2} + {{1 - \cos 2x} \over 2} \cr
& \Leftrightarrow \cos 8x + \cos 6x = \cos 4x + \cos 2x \cr
& \Leftrightarrow 2\cos 7x\cos x = 2\cos 3x\cos x \cr
& \Leftrightarrow 2\cos x\left( {\cos 7x - \cos 3x} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos x = 0 \hfill \cr
\cos 7x = \cos 3x \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over 2} + k\pi \hfill \cr
7x = 3x + k2\pi \hfill \cr
7x = - 3x + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over 2} + k\pi \hfill \cr
x = {{k\pi } \over 2} \hfill \cr
x = {{k\pi } \over 5} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {{k\pi } \over 2} \hfill \cr
x = {{k\pi } \over 5} \hfill \cr} \right.\,\,\,\left( {k \in Z} \right) \cr} \)
