Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\):
\(\left(\sqrt{3}+1\right)\left(tanx+1\right)=1+tan^2x\)
\(\Leftrightarrow tan^2x-\left(\sqrt{3}+1\right)tanx-\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\dfrac{1+\sqrt{3}-\sqrt{4+6\sqrt{3}}}{2}\\tanx=\dfrac{1+\sqrt{3}+\sqrt{4+6\sqrt{3}}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\dfrac{1+\sqrt{3}-\sqrt{4+6\sqrt{3}}}{2}\right)+k\pi\\x=arctan\left(\dfrac{1+\sqrt{3}+\sqrt{4+6\sqrt{3}}}{2}\right)+k\pi\end{matrix}\right.\)
Đúng 0
Bình luận (0)
