Câu hỏi của Julian Edward

Question

giải các pt

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

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

Nguyễn Việt Lâm · Accepted Answer

a/ ĐKXĐ: ...

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

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

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

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

$\Rightarrow\left[{}\begin{matrix}x=k\pi\x=\frac{\pi}{3}+k\pi\end{matrix}\right.$Câu trả lời: a/ ĐKXĐ: ...

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

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

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

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

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

Nguyễn Việt Lâm · Answer

b/ ĐKXĐ: ...

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

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

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

$\Leftrightarrow\frac{1}{cosx}=2$

$\Rightarrow cosx=\frac{1}{2}$

$\Rightarrow x=\pm\frac{\pi}{3}+k2\pi$Câu trả lời: b/ ĐKXĐ: ...

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

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

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

$\Leftrightarrow\frac{1}{cosx}=2$

$\Rightarrow cosx=\frac{1}{2}$

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

Bài 2: Phương trình lượng giác cơ bản