[tex]\displaystyle\\
\sin2x=2\cos^2x~~~\Big| : 2\cos^2x\\\\
\frac{\sin2x}{2\cos^2x}=1\\\\\\
\frac{2\sin x \cos x}{2\cos^2x}=1 \\\\\\
\frac{\sin x\cos x}{\cos^2x}=1\\\\\\
\frac{\sin x}{\cos x}=1\\\\
\text{tg}~x=1\\\\
x=\text{arctg} 1=\boxed{\bf 45^o}=\boxed{\bf \frac{\pi}{4} +k\pi}\\\\\\
\text{Verificare in ecuatia initiala:}\\
\sin2x=2\cos^2x\\\\
\sin2\cdot 45^o=2\cos^245^o\\\\
\sin90^o=2\cos^245^o\\\\
1=2\cdot \left(\frac{ \sqrt{2} }{2} \right)^2 \\\\
1=2\cdot \frac{ 2 }{4}\\\\
1=\frac{4}{4}\\\\
1=1~~~~OK[/tex]
