Question

Giải pt

\(sin^2\frac{x}{2}+sinx-2cos^2\frac{x}{2}=\frac{1}{2}\)

Answer 1

\(\frac{1}{2}-\frac{1}{2}cosx+sinx-1-cosx=\frac{1}{2}\)

\(\Leftrightarrow2sinx-3cosx=2\)

\(\Leftrightarrow\frac{2}{\sqrt{13}}sinx-\frac{3}{\sqrt{13}}cosx=\frac{2}{\sqrt{13}}\)

Đặt \(\frac{2}{\sqrt{13}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Leftrightarrow sinx.cosa-cosx.sina=cosa\)

\(\Leftrightarrow sin\left(x-a\right)=sin\left(\frac{\pi}{2}-a\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-a=\frac{\pi}{2}-a+k2\pi\\x-a=\frac{\pi}{2}+a+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{2}+2a+k2\pi\end{matrix}\right.\)