Đáp án:
\(\left[ \matrix{
x = {\pi \over 4} + k\pi \hfill \cr
x = {1 \over 2}\arcsin {3 \over 5} + k\pi \hfill \cr
x = {\pi \over 2} - {1 \over 2}\arcsin {3 \over 5} + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\eqalign{
& 2010: \cr
& 4\cos {{5x} \over 2}\cos {{3x} \over 2} + 2\left( {8\sin x - 1} \right)\cos x = 5 \cr
& \Leftrightarrow 2\left( {\cos 4x + \cos x} \right) + 16\sin x\cos x - 2\cos x = 5 \cr
& \Leftrightarrow 2\cos 4x + 8\sin 2x = 5 \cr
& \Leftrightarrow 2\left( {1 - 2{{\sin }^2}2x} \right) + 8\sin 2x = 5 \cr
& \Leftrightarrow - 5{\sin ^2}2x + 8\sin 2x - 3 = 0 \cr
& \Leftrightarrow \left[ \matrix{
\sin 2x = 1 \hfill \cr
\sin 2x = {3 \over 5} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
2x = {\pi \over 2} + k2\pi \hfill \cr
2x = \arcsin {3 \over 5} + k2\pi \hfill \cr
2x = \pi - \arcsin {3 \over 5} + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over 4} + k\pi \hfill \cr
x = {1 \over 2}\arcsin {3 \over 5} + k\pi \hfill \cr
x = {\pi \over 2} - {1 \over 2}\arcsin {3 \over 5} + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} \)
