Question

A)tg^x-3tgx+2=0 в)3cos^x-5cosx+2=0 б)2sin^x-3sinx+1=0 г)4ctg^x-6ctgx+2=0

Answer 1

$1)\;tg^2x-3tgx+2=0|\;tgx=t\\t^2-3t+2=0\\D:9-8=1\\t_1,_2=\frac{3\pm1}{2}; \quad t_1=2, \;\;t_2=1;\\\\tgx_1=2\\x_1=arctg2+\pi n, \; n\in Z;\\\\tgx_2=1\\x_2=\frac{\pi}{4}+\pi n, \; n\in Z;\\\\\\ 2)\;3cos^2x-5cosx+2=0|\; cosx=t\\3t^2-5t+2=0\\D:25-24=1\\t_1,_2=\frac{5\pm1}{6}; \quad t_1=1, \;\;t_2=\frac{2}{3};\\\\cosx_1=1\\x_1=2\pi n, \; n\in Z;\\\\cosx_2=\frac{2}{3}\\x_2=\pm arccos\frac{2}{3}+2\pi n, \; n\in Z;$

$3)\;2sin^2x-3sinx+1=0|\; sinx=t\\2t^2-3t+1=0\\D:9-8=1\\t_1,_2=\frac{9\pm1}{4} ;\quad t_1=\frac{5}{4}, \;\;t_2=2;\\\\sinx_1 \neq \frac{5}{4}\\sinx \in [-1;1];\\\\sinx_2 \neq 2\\sinx_2 \in [-1;1];\\\\\\4)\;4ctg^2x-6ctgx+2=0|\;ctgx=t\\4t^2-6t+2=0\\\frac{D}{4}:(\frac{6}{2})^2-2*4=1\\t_1,_2=\frac{3\pm1}{4}; \quad t_1=1, \;\;t_2\frac{1}{2};\\\\ctgx_1=1\\x_1=\frac{\pi}{4}+\pi n, \; n\in Z;\\\\ctgx_2=\frac{1}{2}\\x_2=arcctg\frac{1}{2}+\pi n, \; n\in Z.$