Giải các phương trình lượng giác Sin2x. Cos2x+1/4=0

Đáp án:

 ${\left[\begin{aligned}x=\dfrac{-\pi}{24}+\dfrac{k\pi}{2}\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}{2}\end{aligned}\right.},(k\in \mathbb{Z})\\$

Giải thích các bước giải:

 $\sin2x.\cos2x+\dfrac{1}{4}=0\\
\Leftrightarrow \dfrac{1}{2}\sin4x=\dfrac{-1}{4}\\
\Leftrightarrow \sin4x=\dfrac{-1}{4}.2=\dfrac{-1}{2}\\
\Leftrightarrow {\left[\begin{aligned}4x=\dfrac{-\pi}{6}+k2\pi\\4x=\pi+\dfrac{\pi}{6}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{-\pi}{24}+\dfrac{k\pi}{2}\\4x=\dfrac{7\pi}{6}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{-\pi}{24}+\dfrac{k\pi}{2}\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}{2}\end{aligned}\right.},(k\in \mathbb{Z})\\$