Question

Va roog mult.Rezolvați ecuația sin2x=2cos^2 x.Multumesc muult.

Answer 1

   
[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]