Giải phương trình :) :)

T

trantien.hocmai

$$\cos ^22x+\dfrac{1}{4}.\sin ^24x+1=\sin 4x.\cos 2x+\sin ^2x \\
\leftrightarrow \cos ^22x-\sin 4x.\cos 2x+\dfrac{1}{4}\sin ^24x+1-\sin ^2x=0 \\
\leftrightarrow (\cos 2x-\dfrac{1}{2}\sin 4x)^2+\cos ^2x=0 \\
\leftrightarrow \begin{cases} \cos 2x-\dfrac{1}{2}\sin 4x=0 \\ \cos x=0 \end{cases}$$
 
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