Giải PT 2sin ( x+ $\frac{\pi}{3}$ ) -sin (2x- $\frac{\pi}{6}$ ) = 1/2 15/08/2021 Bởi Adeline 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 $ ⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.\) Bình luận
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 $
⇔ \(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.\)