Phương trình $\left( {\sqrt 3 - 1} \right)\sin x - \left( {\sqrt 3 + 1} \right)\cos x + \sqrt 3 - 1 = 0$ có các nghiệm là
A. $\left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right.$$,k \in \mathbb{Z}$.
B. $\left[ \begin{array}{l}x = - \frac{\pi }{2} + k2\pi \\x = \frac{\pi }{3} + k2\pi \end{array} \right.,k \in \mathbb{Z}$.
C. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{9} + k2\pi \end{array} \right.,k \in \mathbb{Z}$.
D. $\left[ \begin{array}{l}x = - \frac{\pi }{8} + k2\pi \\x = \frac{\pi }{{12}} + k2\pi \end{array} \right.,k \in \mathbb{Z}$.
A. $\left[ \begin{array}{l}x = - \frac{\pi }{4} + k2\pi \\x = \frac{\pi }{6} + k2\pi \end{array} \right.$$,k \in \mathbb{Z}$.
B. $\left[ \begin{array}{l}x = - \frac{\pi }{2} + k2\pi \\x = \frac{\pi }{3} + k2\pi \end{array} \right.,k \in \mathbb{Z}$.
C. $\left[ \begin{array}{l}x = - \frac{\pi }{6} + k2\pi \\x = \frac{\pi }{9} + k2\pi \end{array} \right.,k \in \mathbb{Z}$.
D. $\left[ \begin{array}{l}x = - \frac{\pi }{8} + k2\pi \\x = \frac{\pi }{{12}} + k2\pi \end{array} \right.,k \in \mathbb{Z}$.
Chọn B.
Ta có $\tan \frac{{5\pi }}{{12}} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}$. Chia hai vế PT cho $\sqrt 3 - 1$ được
PT: $\sin x - \tan \frac{{5\pi }}{{12}}.\cos x + 1 = 0$ <=>$\sin x.\cos \frac{{5\pi }}{{12}} - \cos x.\sin \frac{{5\pi }}{{12}} + \cos \frac{{5\pi }}{{12}} = 0$ <=>$\sin \left( {x - \frac{{5\pi }}{{12}}} \right) = - \cos \frac{{5\pi }}{{12}}$ <=>$\sin \left( {x - \frac{{5\pi }}{{12}}} \right) = \sin \left( { - \frac{\pi }{{12}}} \right)$ <=> $\left[ \begin{array}{l}x - \frac{{5\pi }}{{12}} = - \frac{\pi }{{12}} + k2\pi \\x - \frac{{5\pi }}{{12}} = \pi + \frac{\pi }{{12}} + k2\pi \end{array} \right.$ <=>$\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.$ <=>$\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.$$(k \in \mathbb{Z})$
Ta có $\tan \frac{{5\pi }}{{12}} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}$. Chia hai vế PT cho $\sqrt 3 - 1$ được
PT: $\sin x - \tan \frac{{5\pi }}{{12}}.\cos x + 1 = 0$ <=>$\sin x.\cos \frac{{5\pi }}{{12}} - \cos x.\sin \frac{{5\pi }}{{12}} + \cos \frac{{5\pi }}{{12}} = 0$ <=>$\sin \left( {x - \frac{{5\pi }}{{12}}} \right) = - \cos \frac{{5\pi }}{{12}}$ <=>$\sin \left( {x - \frac{{5\pi }}{{12}}} \right) = \sin \left( { - \frac{\pi }{{12}}} \right)$ <=> $\left[ \begin{array}{l}x - \frac{{5\pi }}{{12}} = - \frac{\pi }{{12}} + k2\pi \\x - \frac{{5\pi }}{{12}} = \pi + \frac{\pi }{{12}} + k2\pi \end{array} \right.$ <=>$\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = \frac{{3\pi }}{2} + k2\pi \end{array} \right.$ <=>$\left[ \begin{array}{l}x = \frac{\pi }{3} + k2\pi \\x = - \frac{\pi }{2} + k2\pi \end{array} \right.$$(k \in \mathbb{Z})$
Nguồn: Học Lớp