c) ĐK :$\cos x\neq 0$$\Leftrightarrow \frac{1-2\sin ^2\frac{x }{2}\cos ^2\frac{x }{2}}{1-\sin x}=(1+\sin x)(\tan ^2x+\frac{1}{2})$$\Leftrightarrow 1-\frac{1}{2}\sin ^2x =(1-\sin x)(1+\sin x)(\tan ^2x+\frac{1}{2})$$\Leftrightarrow \cos ^2x(\tan ^2x+\frac{1}{2})-1+\frac{1}{2}\sin ^2x=0$$\Leftrightarrow 3\sin ^2x+\cos ^2x-2=0$$\Leftrightarrow \cos ^2x=\frac{1}{2}$Vậy ......
c) ĐK :$\cos x\neq 0$$\Leftrightarrow \frac{1-2\sin ^2\frac{\pi }{2}\cos ^2\frac{\pi }{2}}{1-\sin x}=(1+\sin x)(\tan ^2x+\frac{1}{2})$$\Leftrightarrow 1-\frac{1}{2}\sin ^2\pi =(1-\sin x)(1+\sin x)(\tan ^2x+\frac{1}{2})$$\Leftrightarrow \cos ^2x(\tan ^2x+\frac{1}{2})-1=0$$\Leftrightarrow 2\sin ^2x+\cos ^2x-2=0$$\Leftrightarrow \cos ^2x=0(ktm)$Vậy ......
c) ĐK :$\cos x\neq 0$$\Leftrightarrow \frac{1-2\sin ^2\frac{
x }{2}\cos ^2\frac{
x }{2}}{1-\sin x}=(1+\sin x)(\tan ^2x+\frac{1}{2})$$\Leftrightarrow 1-\frac{1}{2}\sin ^2
x =(1-\sin x)(1+\sin x)(\tan ^2x+\frac{1}{2})$$\Leftrightarrow \cos ^2x(\tan ^2x+\frac{1}{2})-1
+\frac{1}{2}\sin ^2x=0$$\Leftrightarrow
3\sin ^2x+\cos ^2x-2=0$$\Leftrightarrow \cos ^2x=
\frac{1}{2}$Vậy ......