Question

Побыстрее пожалуйста!!!!!!!!!​

Answer 1

$\displaystyle\bf\\2Sin^{2} x-5Sinx+2=0\\\\Sinx=m \ , \ \ -1\leq m\leq 1\\\\2m^{2} -5m+2=0\\\\D=(-5)^{2} -4\cdot 2\cdot 2=25-16=9=3^{2} \\\\\\m_{1}=\frac{5-3}{4} =\frac{1}{2} \\\\\\m_{2} =\frac{5+3}{4} =2 > 1-neyd\\\\\\Sinx=\frac{1}{2} \\\\\\x=(-1)^{n} arcSin\frac{1}{2} +\pi n,n\in Z\\\\\\\boxed{\boxed{x=(-1)^{n} \frac{\pi }{6} +\pi n,n\in Z}}$

Answer 2

Ответ:    $x=(-1)^{n}\cdot \dfrac{\pi }{6}+\pi n\ ,\ n\in Z\ .$

$2sin^2x-5sinx+2=0$

Сделаем замену:   $t=sinx\ \ ,\ \ |sinx|\leq 1\ \ \Rightarrow \ \ \ 2t^2-5t+2=0\ ,$

$D=b^2-4ac=25-4\cdot 2\cdot 2=9\ \ ,\ \ x_1=\dfrac{5-3}{4}=\dfrac{1}{2}\ ,\ x_1=\dfrac{5+3}{4}=2 > 1\\\\\\sinx=\dfrac{1}{2}\ \ \Rightarrow \ \ \underline{\ x=(-1)^{n}\cdot \dfrac{\pi }{6}+\pi n\ ,\ n\in Z\ }$