Giải các phương trình : 1) cot^2x=1 2) trị tuyệt đối(cos(x))=1/2 3) sin^2(x-pi/4)=cos^2x

1) ${\cot}^2x=1$ Đk: $\sin x\ne 0\Leftrightarrow x\ne k\pi,(k\in\mathbb Z)$ Phương trình $\Rightarrow \dfrac{{\cos}^2x}{{\sin }^2x}=1$ $\Rightarrow {\cos}^2x={\sin}^2x$ $\Rightarrow \left[\begin{array}{l} \cos x=\sin x \\ \cos x=-\sin x \end{array} \right .$ $\Rightarrow \left[\begin{array}{l} \cos x=\cos\left({\dfrac{\pi}{2}-x}\right) \\ \cos x=\sin (-x)=\cos\left({\dfrac{\pi}{2}+x}\right) \end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} x=\pm\left({\dfrac{\pi}{2}-x}\right) +k2\pi\\ \ x=\pm\left({\dfrac{\pi}{2}+x}\right)+k2\pi(tm) \end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} x={\pi}{4}+k\pi\\ \ x=-\dfrac{\pi}{4}k\pi (tm)\end{array} \right .(k\in\mathbb Z)$ 2) $|\cos x|=\dfrac{1}{2}$ $\Rightarrow \left[ \begin{array}{l} \cos x=\dfrac{1}{2} \\ \cos x=-\dfrac{1}{2} \end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} x=\pm\dfrac{\pi}{3}+k2\pi \\ x=\dfrac{2\pi}{3}+k2\pi \end{array} \right .(k\in\mathbb Z)$ 3) ${\sin}^2\left({x-\dfrac{\pi}{4}}\right)={\cos}^2x$ $\Rightarrow \left[\begin{array}{l} \sin\left({x-\dfrac{\pi}{4}}\right)=\cos x \\ \sin\left({x-\dfrac{\pi}{4}}\right)=-\cos x\end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} \sin\left({x-\dfrac{\pi}{4}}\right)=\sin\left({ \dfrac{\pi}{2}-x}\right) \\ \sin\left({x-\dfrac{\pi}{4}}\right)=-\sin\left({ \dfrac{\pi}{2}-x}\right)=\sin\left({ x-\dfrac{\pi}{2}}\right)\end{array} \right .$ $\Rightarrow \left[ \begin{array}{l} \left[ \begin{array}{l} x-\dfrac{\pi}{4}=\left({ \dfrac{\pi}{2}-x}\right)+k2\pi \\ x-\dfrac{\pi}{4}=\pi-\left({ \dfrac{\pi}{2}-x}\right)+k2\pi \end{array} \right . \\\left[ \begin{array}{l} x-\dfrac{\pi}{4}=\left({ x-\dfrac{\pi}{2}}\right)+k2\pi \\ x-\dfrac{\pi}{4}=\pi-\left({ x-\dfrac{\pi}{2}}\right)+k2\pi \end{array} \right . \end{array} \right .$ $\Rightarrow\left[ \begin{array}{l} x=\dfrac{3}{8}+k\pi\\ x=\dfrac{7}{8}+k\pi \end{array} \right .(k\in\mathbb Z)$.