Câu 1: a) $\sin x=\dfrac{1}{2}$
$\Rightarrow \left[\begin{array}{l} x=\dfrac{\pi}{6}+k2\pi \\ x=\dfrac{5\pi}{6}+k2\pi\end{array} \right .(k\in\mathbb Z)$
b) Đk: $\cos x\ne0\Rightarrow x\dfrac{\pi}{2}+k\pi,(k\in\mathbb Z)$
$\tan x=\dfrac{3}{2}$
$\Rightarrow x=\arctan\dfrac{3}{2}+k\pi,(k\in\mathbb Z)$
c) $\cos 5x=-\cos x=\cos(\pi-x)$
$\Rightarrow \left\{ \begin{array}{l} 5x=\pi-x+k\pi\\ 5x=-\pi+x+k\pi\end{array} \right .\Rightarrow \left\{ \begin{array}{l} x=\dfrac{\pi}{6}+k\dfrac{\pi}{6}\\ x=\dfrac{-\pi}{4}+k\dfrac{\pi}{4}\end{array} \right .$
Câu 3: $\dfrac{\sqrt3}{2}\sin x+\dfrac{1}{2}\cos x=1$
Đặt $\dfrac{\sqrt3}{2}=\cos \alpha$ và $\dfrac{1}{2}=\sin \alpha$
$\Rightarrow \cos\alpha\sin x+\sin\alpha\cos x=1$
$\Rightarrow \sin(x+\alpha)=1$
$\Rightarrow x+\alpha=\dfrac{\pi}{2}+k2\pi$
$\Rightarrow x=-\alpha+\dfrac{\pi}{2}+k2\pi,(k\in\mathbb Z)$
Câu 2: a) $2{\cos}^2x+5\cos x+5-2=0$
$\Rightarrow \left[ \begin{array}{l}\cos x=-1(tm) \\ \cos x=-1,5<-1(l) \end{array} \right .$
$\Rightarrow x=\pi+k2\pi,(k\in\mathbb Z)$
b)
