Question

giải phương trình sau :

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

Answer 1

ĐK: \(1+cosx\ne0\Leftrightarrow\cos x\ne-1\Leftrightarrow x\ne\pi+k2\pi\)

Đặt : \(\tan\frac{x}{2}=\alpha\) . Ta có: \(\left\{{}\begin{matrix}\cos x=\frac{1-\alpha^2}{1+\alpha^2}\\\sin x=\frac{2\alpha}{1+\alpha^2}\end{matrix}\right.\)

Ta có phương trình:

\(\frac{1+\frac{2\alpha}{1+\alpha^2}}{1+\frac{1-\alpha^2}{1+\alpha^2}}=\frac{1}{2}\Leftrightarrow\frac{\alpha^2+2\alpha+1}{2}=\frac{1}{2}\)

\(\Leftrightarrow\left(\alpha+1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\alpha=-2\\\alpha=0\end{matrix}\right.\)

Bạn làm tiếp nhé!