Question

Sinx+cosx+sinxcosx=1 решите уравнение?

Answer 1

$sinx+cosx+sinxcosx=1\\\\sinx+cosx=t,\; t^2=sin^2x+cos^2x+2sinxcosx=1+2sinxcosx\; \to \\\\sinxcosx=\frac{1}{2}(t^2-1)\; \; \Rightarrow \\\\ t+\frac{1}{2}(t^2-1)=1\\\\t+\frac{1}{2}t^2-\frac{1}{2}-1=0\, |\cdot 2\\\\t^2+2t-3=0\\\\t_1=-3,\; t_2=1\\\\1)\; sinx+cosx=-3\; \; net\; reshenij,t.k.$

$|sinx| \leq 1,\; |cosx| \leq 1.$

$2) sinx+cosx=1 |:\sqrt 2$

$\frac{1}{\sqrt2}sinx+\frac{1}{\sqrt2}cosx=\frac{1}{\sqrt2}\\\\cos\frac{\pi}{4}sinx+sin\frac{\pi}{4}cosx=\frac{1}{\sqrt2}\\\\sin(x+\frac{\pi}{4})=\frac{1}{\sqrt2}\\\\x+\frac{\pi}{4}=(-1)^{n}\frac{\pi}{4}+\pi n,\; n\in Z\\\\x=(-1)^{n}\frac{\pi}{4}-\frac{\pi}{4}+\pi n,\; n\in Z\\\\x=((-1)^{n}-1)\cdot \frac{\pi}{4}+\pi n,\; n\in Z$