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Решите задачу:

$1)\; 5sin^2x-7sinx\cdot cosx+4cos^2x=1\\\\5sin^2x-7sinx\cdot cosx+4cos^2x=sin^2x+cos^2x\\\\4sin^2x-7sinx\cdot cosx+3cos^2x=0\, |:cos^2x\ne 0\\\\4tg^2x-7tgx+3=0\\\\(tgx)_1=1,\; (tgx)_2=\frac{6}{8}=\frac{3}{4}\\\\x_1=-\frac{\pi}{4}+\pi n,\; n\in Z\\\\x_2=arctg\frac{3}{4}+\pi k,\; k\in Z$

$2)\; 5sin^2x-17sinx\cdot cosx+4cos^2x+4=0\\\\5sin^2x-17sinxcosx+4cos^2x+4(sin^2x+cos^2x)=0\\\\9sin^2x-17sinxcosx+8cos^2x=0\, |:cos^2x\ne 0\\\\9tg^2x-17tgx+8=0\\\\(tgx)_1=1,\; (tgx)_2=\frac{16}{18}=\frac{8}{9}\\\\x_1=\frac{\pi}{4}+\pi n,\; n\in Z\\\\x_2=arctg\frac{8}{9}+\pi k,\; k\in Z$

$3)\; 3cos^2x-sin2x=0,5\\\\3cos^2x-2sinxcosx-0,5(sin^2x+cos^2x)=0\, |\cdot 2\\\\6cos^2x-4sinxcosx-sin^2x-cos^2x=0\, |:cos^2x\ne 0\\\\5tg^2x-4tgx-1=0\\\\(tgx)_1=-\frac{1}{5},\; (tgx)_2=1\\\\x_1=-arctg\frac{1}{5}+\pi n,\; n\in Z\\\\x_2=\frac{\pi}{4}+\pi k,\; k\in Z$