Đáp án:
12)
Đáp án A
13)
Đáp án B
Giải thích các bước giải:
$12)
2\cos2x+1=0\\
\Leftrightarrow \cos2x=\dfrac{-1}{2}\\
\Leftrightarrow {\left[\begin{aligned}2x=\dfrac{2\pi}{3}+k2\pi\\2x=-\dfrac{2\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{\pi}{3}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{aligned}\right.}, (k \in \mathbb{Z})\\
+) 0<x<\dfrac{\pi}{2}\\
\Leftrightarrow 0<\dfrac{\pi}{3}+k\pi<\dfrac{\pi}{2}\\
\Leftrightarrow 0-\dfrac{\pi}{3}<k\pi<\dfrac{\pi}{2}-\dfrac{\pi}{3}\\
\Leftrightarrow -\dfrac{\pi}{3}<k\pi<\dfrac{\pi}{6}\\
\Leftrightarrow -\dfrac{1}{3}<k<\dfrac{1}{6}\\
\Rightarrow k=0\\
\Rightarrow x=\dfrac{\pi}{3}\\
+) 0<x<\dfrac{\pi}{2}\\
\Leftrightarrow 0<-\dfrac{\pi}{3}+k\pi<\dfrac{\pi}{2}\\
\Leftrightarrow 0+\dfrac{\pi}{3}<k\pi<\dfrac{\pi}{2}+\dfrac{\pi}{3}\\
\Leftrightarrow \dfrac{\pi}{3}<k\pi<\dfrac{5\pi}{6}\\
\Leftrightarrow \dfrac{1}{3}<k<\dfrac{5}{6}\\
\Rightarrow k=\varnothing $
Đáp án A
13)
$2\sin3x-\sqrt{3}=0\\
\Leftrightarrow \sin3x=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow {\left[\begin{aligned}3x=\dfrac{\pi}{3}+k2\pi\\3x=\pi-\dfrac{\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\\3x=\dfrac{2\pi}{3}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\\x=\dfrac{2\pi}{9}+\dfrac{k2\pi}{3}\end{aligned}\right.},(k \in \mathbb{Z})\\
+) 0<x<\pi\\
\Leftrightarrow 0<\dfrac{\pi}{9}+\dfrac{k2\pi}{3}<\pi\\
\Leftrightarrow -\dfrac{\pi}{9}<\dfrac{k2\pi}{3}<\pi-\dfrac{\pi}{9}\\
\Leftrightarrow -\dfrac{\pi}{9}<\dfrac{k2\pi}{3}<\dfrac{8\pi}{9}\\
\Leftrightarrow -\dfrac{\pi}{9}.\dfrac{3}{2\pi}<k<\dfrac{8\pi}{9}.\dfrac{3}{2\pi}\\
\Leftrightarrow -\dfrac{1}{6}<k<\dfrac{4}{3}\\
\Rightarrow k=0;1\\
\Rightarrow x=\dfrac{\pi}{9};\dfrac{7\pi}{9}\\
+) 0<x<\pi\\
\Leftrightarrow 0<\dfrac{2\pi}{9}+\dfrac{k2\pi}{3}<\pi\\
\Leftrightarrow -\dfrac{2\pi}{9}<\dfrac{k2\pi}{3}<\pi-\dfrac{2\pi}{9}\\
\Leftrightarrow -\dfrac{2\pi}{9}<\dfrac{k2\pi}{3}<\dfrac{7\pi}{9}\\
\Leftrightarrow -\dfrac{2\pi}{9}.\dfrac{3}{2\pi}<k<\dfrac{7\pi}{9}.\dfrac{3}{2\pi}\\
\Leftrightarrow -\dfrac{1}{3}<k<\dfrac{7}{6}\\
\Rightarrow k=0;1\\
\Rightarrow x=\dfrac{2\pi}{9};\dfrac{8\pi}{9}$
Đáp án B
