Đáp án:
$g)x=\dfrac{5\pi}{6}+k\pi,(k\in \mathbb{Z})\\
k)
x=\dfrac{\pi}{3}+k2\pi,(k\in \mathbb{Z})\\
l)
{\left[\begin{aligned}x=\arccos\dfrac{-1}{5}+k2\pi\\x=-\arccos\dfrac{-1}{5}+k2\pi\end{aligned}\right.},(k\in \mathbb{Z})$
Giải thích các bước giải:
$g)\cos x-\sqrt{3}\sin x=0\\
\Leftrightarrow \dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x=0\\
\Leftrightarrow \cos\dfrac{\pi}{3}\cos x-\sin\dfrac{\pi}{3}\sin x=0\\
\Leftrightarrow \cos\left ( x-\dfrac{\pi}{3} \right )=0\\
\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\\
\Leftrightarrow x=\dfrac{\pi}{2}+\dfrac{\pi}{3}+k\pi\\
\Leftrightarrow x=\dfrac{5\pi}{6}+k\pi,(k\in \mathbb{Z})\\
k)
\sqrt{3}\sin x+\cos x-2=0\\
\Leftrightarrow \dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=\dfrac{2}{2}=1\\
\Leftrightarrow \sin \dfrac{\pi}{3}\sin x+\cos\dfrac{\pi}{3}\cos x=1\\
\Leftrightarrow \cos \left ( x-\dfrac{\pi}{3} \right )=1\\
\Leftrightarrow x-\dfrac{\pi}{3}=k2\pi\\
\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi,(k\in \mathbb{Z})\\
l)
4\sin^2\dfrac{x}{2}+3\cos x-1=0\\
\Leftrightarrow 2.(\cos x+1)+3\cos x-1=0\\
\Leftrightarrow 2\cos x+2+3\cos x-1=0\\
\Leftrightarrow 5\cos x=-1\\
\Leftrightarrow \cos x=\dfrac{-1}{5}\\
\Leftrightarrow {\left[\begin{aligned}x=\arccos\dfrac{-1}{5}+k2\pi\\x=-\arccos\dfrac{-1}{5}+k2\pi\end{aligned}\right.},(k\in \mathbb{Z})\\$
