Question

\(sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)tan^2x-cos^2\dfrac{x}{2}=0\)

Answer 1

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

\(\left[1-cos\left(x-\dfrac{\pi}{2}\right)\right]\dfrac{sin^2x}{cos^2x}-\left(1+cosx\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\dfrac{1-cos^2x}{1-sin^2x}-\left(1+cosx\right)=0\)

\(\Leftrightarrow\dfrac{\left(1-sinx\right)\left(1-cosx\right)\left(1+cosx\right)}{\left(1-sinx\right)\left(1+sinx\right)}-\left(1+cosx\right)=0\)

\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{1-cosx}{1+sinx}-1\right)=0\)

\(\Leftrightarrow\dfrac{\left(1+cosx\right)\left(-cosx-sinx\right)}{1+sinx}=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)