Đáp án:
$x=\dfrac{11\pi}{12}\\
x=\dfrac{7\pi}{12}\\$
Giải thích các bước giải:
$2\sin2x+1=0\\
\Leftrightarrow 2\sin2x=-1\\
\Leftrightarrow \sin2x=\dfrac{-1}{2}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{-\pi}{6}+k2\pi\\2x=\pi-\dfrac{-\pi}{6}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{-\pi}{12}+k\pi\\2x=\dfrac{7\pi}{6}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{-\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{aligned}\right.}\\
+) 0<x<\pi\\
\Leftrightarrow 0<\dfrac{-\pi}{12}+k\pi<\pi\\
\Leftrightarrow \dfrac{\pi}{12}<k\pi<\pi+\dfrac{\pi}{12}\\
\Leftrightarrow \dfrac{\pi}{12}<k\pi<\dfrac{13\pi}{12}\\
\Leftrightarrow \dfrac{1}{12}<k<\dfrac{13}{12}\\
\Rightarrow k=1\\
\Rightarrow x=\dfrac{-\pi}{12}+\pi=\dfrac{11\pi}{12}\\
+) 0<x<\pi\\
\Leftrightarrow 0<\dfrac{7\pi}{12}+k\pi<\pi\\
\Leftrightarrow \dfrac{-7\pi}{12}<k\pi<\pi-\dfrac{7\pi}{12}\\
\Leftrightarrow \dfrac{-7\pi}{12}<k\pi<\dfrac{5\pi}{12}\\
\Leftrightarrow \dfrac{-7}{12}<k<\dfrac{5}{12}\\
\Rightarrow k=0\\
\Rightarrow x=\dfrac{7\pi}{12}\\$
