ĐKXĐ: x/2<>pi/2+kpi
=>x<>pi+k2pi
\(cos\left(x\right)+tan\left(\dfrac{x}{2}\right)=1\)
=>\(1-2\cdot sin^2\left(\dfrac{x}{2}\right)+\dfrac{sin\left(\dfrac{x}{2}\right)}{cos\left(\dfrac{x}{2}\right)}=1\)
\(\Leftrightarrow sin\left(\dfrac{x}{2}\right)\left(-2sin\left(\dfrac{x}{2}\right)+\dfrac{1}{cos\left(\dfrac{x}{2}\right)}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=kpi\\-sinx+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2pi\\x=\dfrac{pi}{2}+k2pi\end{matrix}\right.\)
Đúng 0
Bình luận (0)
