a. Ta có:
\(\eqalign{& \sin \left( {x - {{2\pi } \over 3}} \right) = \cos 2x \cr & \Leftrightarrow \sin \left( {x - {{2\pi } \over 3}} \right) = \sin \left( {{\pi \over 2} - 2x} \right) \cr & \Leftrightarrow \left[ {\matrix{{x - {{2\pi } \over 3} = {\pi \over 2} - 2x + k2\pi } \cr {x - {{2\pi } \over 3} = \pi - {\pi \over 2} + 2x + k2\pi } \cr} } \right. \cr & \Leftrightarrow \left[ {\matrix{{x = {{7\pi } \over {18}} + k{{2\pi } \over 3}} \cr {x = - {{7\pi } \over 6} - k2\pi } \cr} } \right. \cr} \)
b. Với ĐKXĐ của phương trình ta có \(\tan(2x + 45^0) = \cot(45^0 - 2x)\) và \(\tan \left( {180^\circ - {x \over 2}} \right) = \tan \left( { - {x \over 2}} \right)\) nên :
\(\eqalign{
& \tan \left( {2x + 45^\circ } \right)\tan \left( {180^\circ - {x \over 2}} \right) = 1 \cr
& \Leftrightarrow \cot \left( {45^\circ - 2x} \right)\tan \left( { - {x \over 2}} \right) = 1 \cr
& \Leftrightarrow \tan \left( { - {x \over 2}} \right) = \tan \left( {45^\circ - 2x} \right) \cr
& \Leftrightarrow - {x \over 2} = 45^\circ - 2x + k180^\circ \cr
& \Leftrightarrow x = 30^\circ + k120^\circ ,k \in\mathbb Z \cr} \)
c. Ta có:
\(\eqalign{
& \cos 2x - {\sin ^2}x = 0 \cr
& \Leftrightarrow \cos 2x - {{1 - \cos 2x} \over 2} = 0 \cr
& \Leftrightarrow 3\cos 2x - 1 = 0 \Leftrightarrow \cos 2x = {1 \over 3} \cr
& \Leftrightarrow \cos 2x = \cos \alpha \,\left( {\text{ với }\,\cos \alpha = {1 \over 3}} \right) \cr
& \Leftrightarrow x = \pm {\alpha \over 2} + k\pi \,\,(k\in\mathbb Z)\cr} \)
d.
\(\eqalign{& 5\tan x - 2\cot x = 3 \cr & \Leftrightarrow 5\tan x - {2 \over {\tan x}} = 3 \cr & \Leftrightarrow 3{\tan ^2}x - 3\tan x - 2 = 0 \cr & \Leftrightarrow \left[ {\matrix{{\tan x = 1} \cr {\tan x = - {2 \over 5}} \cr} } \right. \Leftrightarrow \left[ {\matrix{{x = {\pi \over 4} + k\pi } \cr {x = \alpha + k\pi } \cr}k\in\mathbb Z } \right. \cr & \text{trong đó}\,\tan \alpha = - {2 \over 5} \cr} \)
