Question

Решите уравнения:
а)
$\sin(2x - \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2}$
б)
$2 \sin^{2}x - 3 \sin(x) + 1 = 0$
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Answer 1

$1)Sin(2x -\frac{\pi }{4})=\frac{\sqrt{2} }{2}\\\\2x-\frac{\pi }{4}=(-1)^{n}arc Sin\frac{\sqrt{2} }{2}+\pi n,n\in Z\\\\2x-\frac{\pi }{4}=(-1)^{n}\frac{\pi }{4}+\pi n,n\in Z\\\\2x=(-1)^{n}\frac{\pi }{4}+\frac{\pi }{4}+\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi }{8}+\frac{\pi }{8}+\frac{\pi n }{2} ,n\in Z$

$2)2Sin^{2}x-3Sinx+1=0\\\\Sinx=m,-1\leq m \leq 1\\\\2m^{2}-3m+1=0\\\\D=(-3)^{2}-4*2*1=9-8=1\\\\m_{1}=\frac{3-1}{4}=\frac{1}{2}\\\\m_{2}=\frac{3+1}{4}=1\\\\1)Sinx=\frac{1}{2} \\\\x=(-1)^{n} arc Sin\frac{1}{2}+\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi }{6}+\pi n,n\in Z\\\\2)Sinx=1\\\\x=\frac{\pi }{2}+2\pi n,n\in Z$