Question

помогите пожалуйста срочно решить ур-е:
cos^2x-sin^2x = (cosx - sinx)^2

Answer 1

$cos^2x-sin^2x=(cosx-sinx)^2\\\\cos2x=cos^2x-2sinx\, cosx+sin^2x\quad (sin^2x+cos^2x=1)\\\\cos2x=1-sin2x\\\\cos2x+sin2x=1\; |:\sqrt2\\\\ \frac{1}{\sqrt2}\, cos2x+ \frac{1}{\sqrt2}\, sin2x= \frac{1}{\sqrt2} \quad (cos \frac{\pi }{4}=sin\frac{\pi }{4}= \frac{1}{\sqrt2}=\frac{\sqrt2}{2})\\\\cos \frac{\pi}{4}\, cos2x+sin \frac{\pi}{4}\, sin2x= \frac{\sqrt2}{2}$

$cos(2x- \frac{\pi}{4})= \frac{\sqrt2}{2} \\\\2x-\frac{\pi}{4}=\pm \frac{\pi }{4}+2\pi n,\; n\in Z\\\\2x= \frac{\pi}{4}\pm \frac{\pi}{4}+2\pi n,\; n\in Z$

$x= \frac{\pi }{8}\pm \frac{\pi }{8}+\pi n= \left [ {{\frac{\pi }{4}+\pi n,\; n\in Z} \atop {\pi n,\; n\in Z\; \; \; \; }} \right.$

Answer 2

сos²x-sin²x=(cosx-sinx)²
cos²x-sin²x-cos²x+2sinxcosx-sin²x=0
2sinxcosx-2sin²x=0
2sinx(cosx-sinx)=0
sinx=0⇒x=πk,k∈z
cosx-sinx=0/cosx
1-tgx=0⇒tgx=1⇒x=π/4+πk,k∈z