Phương trình \(1 + \sin \,x...

B

- Hướng dẫn giải

Ta có: \(1 + \sin x - \cos x - \sin 2x = 0\)

\(\Leftrightarrow {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x + \sin x - \cos x = 0\)

\(\Leftrightarrow {\left( {\sin x - \cos x} \right)^2} + \sin x - \cos x = 0\)

\(\Leftrightarrow \left( {\sin x - \cos x} \right)\left( {\sin x - \cos x + 1} \right) = 0\)

\(\Leftrightarrow \left[ \begin{array}{l}\sin x = \cos x\\\sin x - \cos x = - 1\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}\tan x = 1\\\sin \left( {x - \dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}\end{array} \right.\)

\(\Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x - \frac{\pi }{4} = - \frac{\pi }{4} + k2\pi \\x - \frac{\pi }{4} = \frac{{5\pi }}{4} + k2\pi\end{array} \right.\)

\(\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\pi \\x = k2\pi \\x = \dfrac{{3\pi }}{2} + k2\pi \end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)