Nghiệm của phương trình $\sin x + \sqrt 3 \cos x = \sqrt 2 $ là
A. $x = - \frac{\pi }{4} + k2\pi ,\,x = \frac{{3\pi }}{4} + k2\pi ,k \in \mathbb{Z}$.
B. $x = - \frac{\pi }{{12}} + k2\pi ,\,x = \frac{{5\pi }}{{12}} + k2\pi ,k \in \mathbb{Z}$.
C. $x = \frac{\pi }{3} + k2\pi ,\,x = \frac{{2\pi }}{3} + k2\pi ,k \in \mathbb{Z}$.
D. $x = - \frac{\pi }{4} + k2\pi ,\,x = - \frac{{5\pi }}{4} + k2\pi ,k \in \mathbb{Z}$.
A. $x = - \frac{\pi }{4} + k2\pi ,\,x = \frac{{3\pi }}{4} + k2\pi ,k \in \mathbb{Z}$.
B. $x = - \frac{\pi }{{12}} + k2\pi ,\,x = \frac{{5\pi }}{{12}} + k2\pi ,k \in \mathbb{Z}$.
C. $x = \frac{\pi }{3} + k2\pi ,\,x = \frac{{2\pi }}{3} + k2\pi ,k \in \mathbb{Z}$.
D. $x = - \frac{\pi }{4} + k2\pi ,\,x = - \frac{{5\pi }}{4} + k2\pi ,k \in \mathbb{Z}$.
Chọn B.
Chia hai vế PT cho 2 ta được $\frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = \frac{{\sqrt 2 }}{2}$ <=>$\sin \left( {x + \frac{\pi }{3}} \right) = \sin \frac{\pi }{4}$
<=>$\left[ \begin{array}{l}x + \frac{\pi }{3} = \frac{\pi }{4} + k2\pi \\x + \frac{\pi }{3} = \pi - \frac{\pi }{4} + k2\pi \end{array} \right.$ <=>$\left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k2\pi \\x = \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.$ $(k \in \mathbb{Z})$
Chia hai vế PT cho 2 ta được $\frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = \frac{{\sqrt 2 }}{2}$ <=>$\sin \left( {x + \frac{\pi }{3}} \right) = \sin \frac{\pi }{4}$
<=>$\left[ \begin{array}{l}x + \frac{\pi }{3} = \frac{\pi }{4} + k2\pi \\x + \frac{\pi }{3} = \pi - \frac{\pi }{4} + k2\pi \end{array} \right.$ <=>$\left[ \begin{array}{l}x = - \frac{\pi }{{12}} + k2\pi \\x = \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.$ $(k \in \mathbb{Z})$
Nguồn: Học Lớp