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Question 1

Question 2

Решите задачу:

$1.\;3sin^2x-10sinx+7=0, \quad sinx=t\\3t^2-10t+7=0\\\frac{D}{4}:(\frac{10}{2})^2-7*3=4\\t_1,_2=\frac{5\pm2}{3}, \quad t_2=\frac{7}{3}, \; \; t_2=1;\\\\sinx_1 \neq \frac{7}{3},\quad sinx\in [-1;+1];\\\\sinx_2=1\\x=\frac{\pi}{2}+2\pi n, \; n\in Z;\\\\2.\;8sin^2x+10cosx-1=0\\8(1-cos^2x)+10cosx-1=0|*(-1)\\cos^2x-10cosx-7=0, \quad cosx=t\\t^2-10t-7=0\\\frac{D}{4}:(\frac{10}{2})^2+7=32\\t_1,_2=10\pm4\sqrt2, \quad t_1=10+4\sqrt2\ \textgreater \ 1, \; \; t_2=10-\sqrt2\ \textgreater \ 1, \\cosx\in[-1;1];$

$3.\;4sin^2x+13sinxcosx+10cos^2x=0|:cos^2x,\\cosx \neq 0,\;x \neq \frac{\pi}{2}+\pi n,\;n\in Z;\\4tg^2x+13tgx+10=0, \quad tgx=t\\4t^2+13t+10=0\\D:169-160=9\\t_1,_2=\frac{-13\pm3}{8}, \quad t_1=-\frac{5}{4}, \; t_2=-2;\\tgx_1=-\frac{5}{4}\\x_1=-arctg\frac{5}{4}+\pi n, \; n\in Z;\\\\tgx_2=-2\\x_2=-arctg2+\pi n, \; n\in Z;$

$4.\;3tgx-3ctgx+8=0\\3tgx-\frac{3}{tgx}+8=0|*tgx,\; tgx \neq 0,\;x \neq \pi n, \; n\in Z;\\3tg^2x+8tgx-3=0, \quad tgx=t\\3t^2+8t-3=0\\\frac{D}{4}:(\frac{8}{2})^2+3*3=25\\t_1,_2=\frac{-4\pm5}{3}, \quad t_1=\frac{1}{3}, \; t_2=-3;\\\\tgx_1=\frac{1}{3}\\x_1=arctg\frac{1}{3}+\pi n, \; n\in Z;\\\\tgx_2=-3\\x_2=-arctg3+\pi n, \; n\in Z.$

$5.\;sin2x+4cos^2x=1\\2sinxcosx+4cos^2x-(sin^2x+cos^2x)=0\\3cos^2x+2sinxcosx-sin^2x=0|:sin^2x,\;sinx \neq 0,\;x \neq \pi n,\;n\in Z;\\3ctg^2x+2ctgx-1=0, \quad ctgx=t\\3t^2+2t-1=0\\\frac{D}{2}:(\frac{2}{2})^2+3*1=4;\\t_1,_2=\frac{-1\pm2}{3},\quad t_1=\frac{1}{3}, \; t_2=-1;\\\\ctgx_1=\frac{1}{3}\\x_1=arcctg\frac{1}{3};\\\\ctgx_2=-1\\x_2=\frac{3\pi}{4}+\pi n, \; n\in Z;$

$6.\; 10cos^2x-9sin2x=4cos2x-4\\10cos^2x-9sin2x-4(cos2x-1)=0\\10cos^2x-9*2sinxcosx-4(-2sin^2x)=0\\10cos^2x-18sinxcosx+8sin^2x=0|:2sin^2x\\sinx \neq 0,\;x \neq \pi n,\;n\in Z;\\cos^2x-9sinxcosx=0\\4tg^2x-9tgx+5=0, \quad tgx=t;\\D:81-80=1\\t_1,_2=\frac{9\pm1}{8}, \quad t_1=\frac{5}{4},\;t_2=1;\\\\tgx_1=\frac{5}{4}\\x_1=arctg\frac{5}{4}+\pi n, \; n\in Z;\\\\tgx_2=1\\x_2=\frac{\pi}{4}+\pi n, \; n\in Z.$