Giải phương trình 2sinxcos2x – 1 + 2cos2x – sinx = 0.
Giải phương trình 2sinxcos2x – 1 + 2cos2x – sinx = 0.
Lời giải:
2sinxcos2x – 1 + 2cos2x – sinx = 0
⇔ (2sinxcos2x + 2cos2x) – (1 + sin x) = 0
⇔ 2cos2x(sinx + 1) – (1 + sinx) = 0
⇔ (sinx + 1)(2cos2x – 1) = 0
\[ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\sin x = - 1}\\{\cos 2x = \frac{1}{2}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \frac{\pi }{2} + k2\pi }\\{\begin{array}{*{20}{c}}{x = \frac{\pi }{6} + k\pi }\\{x = - \frac{\pi }{6} + k\pi }\end{array}}\end{array}} \right.\left( {k \in \mathbb{Z}} \right)\].