Giải pt:
$(1+\sqrt3)\sin x+(1-\sqrt3)\cos x=2$
Nếu đặt $\tan \dfrac{x}{2} = t$  (tự xử điều kiện)

PT $\Leftrightarrow (1+\sqrt 3) \dfrac{2t}{1+t^2} +(1-\sqrt 3). \dfrac{1-t^2}{1+t^2}=2$

Pt bậc 2 thôi :v
Ta có:
      $(1+\sqrt3)\sin x+(1-\sqrt3)\cos x=2$
$\Leftrightarrow \dfrac{1+\sqrt3}{2\sqrt2}\sin x-\dfrac{\sqrt3-1}{2\sqrt2}\cos x=\dfrac{1}{\sqrt2}$
$\Leftrightarrow \cos\dfrac{\pi}{12}\sin x-\sin\dfrac{\pi}{12}\cos x=\sin\dfrac{\pi}{4}$
$\Leftrightarrow \sin(x-\dfrac{\pi}{12})=\sin\dfrac{\pi}{4}$
$\Leftrightarrow \left[\begin{array}{l}x-\dfrac{\pi}{12}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{12}=\pi-\dfrac{\pi}{4}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
$\Leftrightarrow \left[\begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{array}\right.,k\in\mathbb{Z}$
ta sẽ chia cả hai vế cho $\sqrt{(1 + \sqrt{3}) ^{2} + (1 - \sqrt{3}) ^{2}} = 2\sqrt{2}$
Đặt $\frac{1 + \sqrt{3} }{2\sqrt{2} }=\cos a , \frac{1 - \sqrt{3}}{2\sqrt{2}}=\sin a$
=>$\cos a\sin x +\sin a\cos x =\frac{1}{\sqrt{2}} => \sin ( a + x) =\sin \frac{\pi }{4}$
=>$\left[ {} \right.\begin{matrix} x= \frac{\pi }{4} - a + k2\pi \\ x= \frac{3\pi }{4} + k2\pi \end{matrix}$

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