Giải phương trình: \({\sin ^2}x = {\cos ^2}x.\)
Giải chi tiết:
Ta có: \({\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\cos ^2}x = 1 - {\sin ^2}x\)
\(\begin{array}{l} \Rightarrow {\sin ^2}x = {\cos ^2}x \Leftrightarrow {\sin ^2}x = 1 - {\sin ^2}x \Leftrightarrow {\sin ^2}x = \frac{1}{2} \Leftrightarrow \sin x = \pm \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \left[ \begin{array}{l}\sin x = \sin \frac{\pi }{4}\\\sin x = \sin \left( { - \frac{\pi }{4}} \right)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = \frac{{3\pi }}{4} + k2\pi \\x = - \frac{\pi }{4} + k2\pi \\x = \frac{{5\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\;\;\left( {k \in Z} \right).\end{array}\)
Chọn B.