Giải PT 2sin ( x+ $\frac{\pi}{3}$ ) -sin (2x- $\frac{\pi}{6}$ ) = 1/2

Đáp án:

\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.\) 

Giải thích các bước giải:

phương trình ⇔ $\cos 2x .\dfrac{1}{2}- \sin 2x .\dfrac{\sqrt{3}}{2}+2\sin x .\dfrac{1}{2} +2\cos x.\dfrac{\sqrt{3}}{2}=\dfrac{1}{2} $ 

⇔ $\cos 2x – \sin 2x .\sqrt{3}+2\sin x +2\cos x\sqrt{3}=1 $ 

⇔ $1-2\sin^2 x+2\sin x – 2\sin x\cos x \sqrt{3}+2\cos x\sqrt{3}=1 $ 

⇔ $-2\sin x(\sin x-1) – 2\sqrt{3}\cos x(\sin x -1)=0 $ 

⇔ $(\sin x-1)(\sin x+\sqrt{3}\cos x )=0 $ ⇔ $(\sin x-1)\sin (x+\dfrac{\pi}{3})=0 $

⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.\)