Question

Срочно!Решите уравнение 5sin^2x-4sin(3n/2+x)=4

Answer 1

$\displaystyle\bf\\5Sin^{2} x-4Sin\Big(\frac{3\pi }{2} +x\Big)=4\\\\5Sin^{2} x+4Cosx-4=0\\\\5\cdot(1-Cos^{2} x)+4Cosx-4=0\\\\5-5Cos^{2} x+4Cosx-4=0\\\\5Cos^{2} x-4Cosx-1=0\\\\Cosx=m \ \ ; \ \ -1\leq m\leq 1\\\\5m^{2} -4m-1=0\\\\D=(-4)^{2} -4\cdot 5\cdot(-1)=16+20=36=6^{2} \\\\\\m_{1} =\frac{4-6}{10} =-\frac{1}{5} \\\\\\m_{2} =\frac{4+6}{10} =1$

$\displaystyle\bf\\1)\\\\Cosx=-\frac{1}{5} \\\\\\x=\pm \ arcCos\Big(-\frac{1}{5} \Big)+2\pi n,n\in Z\\\\2)\\\\Cosx=1\\\\x=2\pi n,n\in Z\\\\\\Otvet \ : \ \pm \ arcCos\Big(-\frac{1}{5} \Big)+2\pi n,n\in Z \ \ ; \ \ 2\pi n,n\in Z$