Đáp án:
\(\left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} = k2\pi \\
x = \frac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.(k \in Z)\)
Giải thích các bước giải:
sin2x+cos2x+3sinx-cosx-2=0
2sinx.cosx-cosx+1-2sin²x+3sinx-2=0
cosx(2sinx-1)-(sinx-1)(2sinx-1)=0
(2sinx-1)(cosx-sinx+1)=0
(2sinx-1)[√2sin(x-$\frac{\pi}{4}$ )-1]=0
\(\left[ \begin{array}{l}
\sin x = \frac{1}{2}\\
\sin (x - \frac{\pi }{4}) = \frac{1}{{\sqrt 2 }}
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} = k2\pi \\
x - \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\
x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} = k2\pi \\
x = \frac{\pi }{2} + k2\pi \\
x = \pi + k2\pi
\end{array} \right.(k \in Z)\)
