Ответ:
Объяснение:
tg^2(x)-4tg(x)+3=0\\x\neq \frac{\pi }{2}+\pi k\\tg(x)=t=>t^2-4t+3=0=>t=1;t=3=>tg(x)=1=>x=\frac{\pi }{4}+ +\pi k\\tg(x)=3=>arctg(3)+\pik\\2)sin(5x)sin(4x)+cos(6x)cos(3x)=0<=>\frac{cos(x)-cos(9x)}{2}+\frac{cos(3x)+cos(9x)}{2}=0\\\frac{cos(x)+cos(3x)}{2}=0<=>cos(2x)cos(x)=0=>x=\frac{\pi }{2}+\pi k\\ x=\frac{\pi }{4}+\frac{\pi k}{2} \\3)2sin^2(x)+3sin(x)cos(x)-3cos^2(x)=0<=>2tg^2(x)+3tg(x)-2=0\\x\neq \frac{\pi }{2}+\pi k\\2t^2+3t-2=0=>t=-2;t=\frac{1}{2}" alt="1)sin^2(x)-4sin(x)cos(x)+3cos^2(x)=0<=>tg^2(x)-4tg(x)+3=0\\x\neq \frac{\pi }{2}+\pi k\\tg(x)=t=>t^2-4t+3=0=>t=1;t=3=>tg(x)=1=>x=\frac{\pi }{4}+ +\pi k\\tg(x)=3=>arctg(3)+\pik\\2)sin(5x)sin(4x)+cos(6x)cos(3x)=0<=>\frac{cos(x)-cos(9x)}{2}+\frac{cos(3x)+cos(9x)}{2}=0\\\frac{cos(x)+cos(3x)}{2}=0<=>cos(2x)cos(x)=0=>x=\frac{\pi }{2}+\pi k\\ x=\frac{\pi }{4}+\frac{\pi k}{2} \\3)2sin^2(x)+3sin(x)cos(x)-3cos^2(x)=0<=>2tg^2(x)+3tg(x)-2=0\\x\neq \frac{\pi }{2}+\pi k\\2t^2+3t-2=0=>t=-2;t=\frac{1}{2}" align="absmiddle" class="latex-formula">
x=-arctg(2)+\pi k\\tg(x)=\frac{1}{2} =>x=arctg(\frac{1}{2} )+\pi k\\4)cos(6x)cos(12x)=cos(8x)cos(10x)<=>0,5(cos(6x)+cos(18x))=0,5(cos(2x)+cos(18x))\\cos(6x)=cos(2x)<=>sin(4x)sin(2x)=0=>x=\frac{\pi k}{4}" alt="tg(x)=-2=>x=-arctg(2)+\pi k\\tg(x)=\frac{1}{2} =>x=arctg(\frac{1}{2} )+\pi k\\4)cos(6x)cos(12x)=cos(8x)cos(10x)<=>0,5(cos(6x)+cos(18x))=0,5(cos(2x)+cos(18x))\\cos(6x)=cos(2x)<=>sin(4x)sin(2x)=0=>x=\frac{\pi k}{4}" align="absmiddle" class="latex-formula">