Question

giải các pt

a) \(\sqrt{3}sinx+cosx=\frac{1}{cosx}\)

b) \(-\frac{1}{2}tan^2x+\frac{2}{cosx}-\frac{5}{2}=0\)

Answer 1

a/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{3}\frac{sinx}{cosx}+1=\frac{1}{cos^2x}\)

\(\Leftrightarrow\sqrt{3}tanx+1=1+tan^2x\)

\(\Leftrightarrow tanx\left(tanx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=0\\tanx=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)

Answer 2

b/ ĐKXĐ: ...

\(\Leftrightarrow tan^2x+1-\frac{4}{cosx}+4=0\)

\(\Leftrightarrow\frac{1}{cos^2x}-\frac{4}{cosx}+4=0\)

\(\Leftrightarrow\left(\frac{1}{cosx}-2\right)^2=0\)

\(\Leftrightarrow\frac{1}{cosx}=2\)

\(\Rightarrow cosx=\frac{1}{2}\)

\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)