Cho phương trình lượng giác 2 cos ( x + π /4 ) − √ 3 = 0 ( ∗ ) . Khi đó

a) Đ, b) S, c) Đ, d) S

a) Ta có \(2\cos \left( {x + \frac{\pi }{4}} \right) - \sqrt 3 = 0 \Leftrightarrow \cos \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2} \Leftrightarrow \cos \left( {x + \frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{6}} \right)\).

b) Ta có \(2\cos \left( {x + \frac{\pi }{4}} \right) - \sqrt 3 = 0 \Leftrightarrow \cos \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2} \Leftrightarrow \cos \left( {x + \frac{\pi }{4}} \right) = \cos \left( {\frac{\pi }{6}} \right)\)

\( \Leftrightarrow x + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi ,k \in \mathbb{Z}\) hoặc \(x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi ,k \in \mathbb{Z}\)

\( \Leftrightarrow x = - \frac{\pi }{{12}} + k2\pi ,k \in \mathbb{Z}\) hoặc \[x = - \frac{{5\pi }}{{12}} + k2\pi ,k \in \mathbb{Z}\].

c)

- Với \(x = - \frac{\pi }{{12}} + k2\pi ,k \in \mathbb{Z}\)

Vì \(x \in \left( { - \pi ;\pi } \right)\) nên \( - \pi < - \frac{\pi }{{12}} + k2\pi < \pi \Leftrightarrow - \frac{{11\pi }}{{12}} < k2\pi < \frac{{13\pi }}{{12}} \Leftrightarrow - \frac{{11}}{{24}} < k < \frac{{13}}{{24}}\)

Do \(k \in \mathbb{Z}\) nên \(k = 0\). Suy ra \({x_1} = - \frac{\pi }{{12}}\)

- Với \[x = - \frac{{5\pi }}{{12}} + k2\pi ,k \in \mathbb{Z}\]

Vì \(x \in \left( { - \pi ;\pi } \right)\) nên \( - \pi < - \frac{{5\pi }}{{12}} + k2\pi < \pi \Leftrightarrow - \frac{{7\pi }}{{12}} < k2\pi < \frac{{17\pi }}{{12}} \Leftrightarrow - \frac{7}{{24}} < k < \frac{{17}}{{24}}\)

Do \(k \in \mathbb{Z}\) nên \(k = 0\). Suy ra \({x_2} = - \frac{{5\pi }}{{12}}\).

d) Ta có \(S = {x_1} + {x_2} = - \frac{{5\pi }}{{12}} - \frac{\pi }{{12}} = - \frac{\pi }{2}\).