Đáp án:
m.
$ x=\dfrac{\pi}{2}+k2\pi$
hoặc $x=\dfrac{-\pi}{6}+k2\pi$
Giải thích các bước giải:
m.$\sqrt[]{3}sinx+cosx=1$
$\rightarrow \dfrac{\sqrt[]{3}}{2}sinx+\dfrac{1}{2}.cosx=\dfrac{1}{2}$
$\rightarrow cos\dfrac{\pi}{6}.sinx+ sin\dfrac{\pi}{6}.cosx=\dfrac{1}{2}$
$\rightarrow cos(x-\dfrac{\pi}{6})=\dfrac{1}{2}$
$\rightarrow x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi$
hoặc $x-\dfrac{\pi}{6}=\dfrac{-\pi}{3}+k2\pi$
$\rightarrow x=\dfrac{\pi}{2}+k2\pi$
hoặc $x=\dfrac{-\pi}{6}+k2\pi$
