Question

Cos²x=0.5tgx*ctgx+sin²x sin3x*cos10x=sin13x tg(x+540градусов)=tg(270 градусов-x) )

Answer 1

$1)cos^2x= \frac{1}{2}tgx*ctgx+sin^2x\\(tgx*ctgx=1)\\cos^2x- \frac{1}{2}*1+sin^2x=0\\cos^2x- \frac{1}{2}+(1-cos^2x)=0\\cos^2x- \frac{1}{2}+1-cos^2x=0\\0.5...\\\\2)cos^2x= \frac{1}{2}tgx*ctgx+sin^2x\\cos^2x=\frac{1}{2} \frac{sinx}{cosx}* \frac{cosx}{sinx}+sin^2x\\cos^2x=\frac{1}{2} \frac{cosxsinx}{cosxsinx}+sin^2x\\cos^2x-\frac{1}{2}+sin^2x\\(cos^2x+sin^2x=1)\\-\frac{1}{2}+1=0.5$

$sin3x*cos10x=sin13x\\\\ \frac{1}{2}sin(3x+10x)+sin(3x-10x)=sin13x\\ \\\frac{1}{2}sin13x+sin(-7x)=sin13x\\\\\frac{1}{2}sin13x-sin13x+sin(-7x)=0\\\\-\frac{1}{2}sin13x+sin(-7x)=0\\\\-\frac{1}{2}(2sin \frac{13x+(-7x)}{2}*cos \frac{13x-(-7x)}{2})=0\\\\-\frac{1}{2}(2sin3x*cos10x)=0\\\\-sin3x*co10x=0\\\\-sin3x=0\\sin3x=0\\3x= \pi n,...\\x= \frac{ \pi }{3}n,...\\\\cos10x=0\\10x= \frac{ \pi }{2}+ \pi n,...\\x= \frac{ \pi }{20}+ \frac{ \pi }{10}n,...$

$tg(x+540)=tg(270-x)\\tg(540+x)=tg(270-x)\\540=3 \pi \\270= \frac{3 \pi }{2}\\tg(3 \pi +x)=tg( \frac{3 \pi }{2}-x)\\tgx=ctgx\\tgx-ctgx=0\\ \frac{sinx}{cosx}- \frac{cosx}{sinx}=0\\sin^2x-cos^2x=0\\(1-cos^2x)-cos^2x=0\\1-cos^2x-co^2x=0\\1-2cos^2x=0\\-2cos^2x=-1\\cos^2x= \frac{1}{2}\\cosx=б\frac{1}{ \sqrt{2}}\\cosx=б\frac{ \sqrt{2}}{2}\\x=б \frac{ \pi }{4}+2 \pi n,...$

Там, где многоточие, там, n∈z