Question

giải các pt

a) \(\frac{1}{sin^2x}=cotx+3\)

b) \(\frac{\sqrt{3}}{sin^2x}=3cotx+\sqrt{3}\)

c) \(9-13cosx+\frac{4}{1+tan^2x}=0\)

d) \(2tan^2x+3=\frac{3}{cosx}\)

Answer 1

a/

ĐKXĐ: ..

\(\Leftrightarrow1+cot^2x=cotx+3\)

\(\Leftrightarrow cot^2x-cotx-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot\left(2\right)+k\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow\sqrt{3}\left(1+cot^2x\right)=3cotx+\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}cot^2x-3cotx=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Answer 2

c/

ĐKXĐ: ...

\(\Leftrightarrow9-13cosx+4.cos^2x=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(4cosx-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}>1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=k2\pi\)

d/

\(\Leftrightarrow2\left(tan^2x+1\right)+1=\frac{3}{cosx}\)

\(\Leftrightarrow\frac{2}{cos^2x}-\frac{3}{cosx}+1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=1\\\frac{2}{cosx}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=k2\pi\)