Question

\(\sin x\left(1+\tan x.\tan\frac{x}{2}\right)+\tan x+2\sqrt{3}=\frac{\sqrt{3}}{\cos^2x}\)

Answer 1

ĐK:\(\begin{cases}cosx\ne0\\cos\frac{x}{2}\ne0\end{cases}\)

\(pt\Leftrightarrow sinx.\left(\frac{cosx.cos\frac{x}{2}+sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}}\right)+tanx+\)\(2\sqrt{3}=\frac{\sqrt{3}}{cos^2x}\)

\(\Leftrightarrow sinx.\frac{cos\frac{x}{2}}{cosx.cos\frac{x}{2}}+tanx+2\sqrt{3}=\sqrt{3}\left(1+tan^2x\right)\)

\(\Leftrightarrow2tanx+2\sqrt{3}=\sqrt{3}\left(1+tan^2x\right)\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}tanx=\sqrt{3}\\tanx=-\frac{1}{\sqrt{3}}\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{cases}\left(k\in Z\right)\left(\frac{t}{m}đk\right)}\)