Question

[tex]sin2x=2cos^2x [\tex]
Determinati solutiile din [0,2pi]

Answer 1

   
[tex]\displaystyle\\ \sin2x=2\cos^2x\\\\ 2\sin x\cos x=2\cos^2x~~\Big|:2\cos x\\\\ \sin x=\cos x~~\Big|:\cos x\\\\ \frac{\sin x}{\cos x}=1\\\\ \text{tg }x=1\\\\ x=\text{arctg}~(1)=\frac{\pi}{4}+k\pi\\\\ \text{In intervalul:}~[0,~2\pi]~\text{tangenta este pozitiva in cadranele I si III.}\\ \text{Rezulta ca avem 2 solutii:}\\\\ \text{Solutia 1 (in cadranul I):}\\\\ \boxed{x_1=\frac{\pi}{4}=45^o}\\\\ \text{Solutia 2 (in cadranul III):}\\\\\boxed{x_2=\frac{\pi}{4}+\pi=180^o+45^o= \frac{5\pi}{4}=225^o}\\\\[/tex]