[tex]sinx+cosx = sinx\cdot cosx|()^2 \Rightarrow \\ \Rightarrow sin^2 x+ cos^2 x+ 2sinx\cdot cosx= sin^2 x \cdot cos^2 x \Rightarrow \\ \\ \Rightarrow 1+ sin2x = \frac{4sin^2x\cdot cos^2 x}{4} \Rightarrow 1+ sin2x = \frac{(sin2x)^2 }{4} \Rightarrow \\ \Rightarrow 4+4sin2x = (sin2x)^2 \\ \\ $ \ Notam sin2x = t \Rightarrow \\ \Rightarrow 4+4t = t^2 \Rightarrow t^2 - 4t - 4 = 0 \\ \\ \Delta = 16+16 = 32 \\ \\ t_{1,2}= \frac{-4\pm 4 \sqrt{2} }{2} \Rightarrow t_{1,2} = \-2\pm 2 \sqrt{2} [/tex]
[tex]\fbox{1} \quad sin2x = 2-2 \sqrt{2} \Rightarrow x = \{(-1)^k\cdot \frac{ arcsin(2-2 \sqrt{2})}{2} +\frac{k\pi}{2}\}\\ \\ \fbox{2} \quad sin2x = 2+2 \sqrt{2} \Rightarrow x = \{(-1)^k\cdot \frac{ arcsin(2+2 \sqrt{2})}{2} +\ \frac{k\pi}{2}\}[/tex]
[tex]k \in \mathbb_{Z}[/tex]
[tex]$ \ $ \ Dar sinx \in [-1,1] \Rightarrow \fbox{1} $ \ este singura solutie adevarata.[/tex]
