Question

решите тригонометрические уравнения
1)2cos^2(3π+x)-5sin(5π/2+x)+2=0
2)1-cos2x+sin2x/1+sin2x+cos2x=√3

Answer 1

$2)\; \; \frac{1-cos2x+sin2x}{1+cos2x+sin2x}=\sqrt3\\\\\frac{2sin^2x+2sinxcosx}{2cos^2x+2sinxcosx}=\sqrt3\\\\\frac{2sinx(sinx+cosx)}{2cosx(cosx+sinx)}=\sqrt3\\\\tgx=\sqrt3\\\\x=\frac{\pi}{3}+\pi n,\; n\in Z$

$1)\; \; 2cos^2(3\pi +x)-5sin(\frac{5\pi }{2}+x)+2=0\\\\\{cos(3\pi +x)=cos(\pi +x)=-cosx,\; \; cos^2(3\pi +x)=(-cosx)^2=cos^2x\\\\sin(\frac{5\pi}{2}+x)=sin(\frac{\pi}{2}+x)=cosx\}\\\\2cos^2x-5cosx+2=0\\\\D=25-16=9\\\\(cosx)_1=\frac{1}{2}\; ,\; x=\pm \frac{\pi}{3}+2\pi n,\; n\in Z\\\\(cosx)_2=2\; \; net\; \; reshenij,\; t.k.\; -1 \leq cosx \leq 1$