Question

1)3(cos x-sin x)=1+cos2x-sin2x 2)2 cos 2x+2cos x sin^2x=cos x

Answer 1

$1)\; \; 3(cosx-sinx)=1+\underbrace {cos2x}_{2cos^2x-1}-\underbrace {sin2x}_{2sinx\, cosx}\\\\3(cosx-sinx)=2cos^2x-2sinx\cdot cosx\\\\3(cosx-sinx)=2cosx(cosx-sinx)\\\\(cosx-sinx)(3-2cosx)=0\\\\a)\; \; cosx-sinx=0\; |:cosx\ne 0\\\\tgx-1=0\; ,\; \; tgx=1\; ,\; \; \underline {x=\frac{\pi}{4}+\pi n,\; n\in Z}\\\\b)\; \; 3-2cosx=0\; ,\; \; cosx=\frac{3}{2}\ \textgreater \ 1\; \; net\; reshenij\\\\\underline {x\in \varnothing }\\\\Otvet:\; \; x= \frac{\pi }{4}+\pi n,\; n\in Z$

$2)\; \; 2cos2x+2cosx\cdot sin^2x=cosx\\\\\star \; cos2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=2cos^2x-1\; \star \\\\\star \; sin^2x+cos^2x=1\; \; \to \; \; sin^2x=1-cos^2x\; \star \\\\2(2cos^2x-1)+2cosx(1-cos^2x)=cosx\\\\4cos^2x-2+2cosx-2cos^3x-cosx=0\\\\4cos^2x-2cos^3x+cosx-2=0\\\\2cos^2x(2-cosx)-(2-cosx)=0\\\\(2-cosx)(2cos^2x-1)=0$

$a)\; \; 2-cosx=0\; ,\; \; cosx=2\ \textgreater \ 1\; \; net\; reshenij\\\\\underline {x\in \varnothing }\\\\b)\; \; \underbrace {2cos^2x-1}_{cos2x}=0\; ,\; \; cos2x=0\; ,\; \; 2x= \frac{\pi }{2}+\pi n\; ,\; n\in Z$

$\underline {x= \frac{\pi }{4}+\frac{\pi n}{2}\; ,\; n\in Z} \; \; -\; \; otvet\; .$