Question

помогите решить тригометрическе уравнение sin(−3π.2+8x)=√2/2 и выбери корни из отрезка [π2;π].

Answer 1

Решение задания прилагаю

Answer 2

$Sin(-\frac{3\pi }{2}+8x)=\frac{\sqrt{2} }{2} \\\\Sin(\frac{3\pi }{2}-8x)=-\frac{\sqrt{2} }{2}\\\\-Cos8x=-\frac{\sqrt{2} }{2}\\\\Cos8x=\frac{\sqrt{2} }{2}\\\\8 \frac{x}{y} x=\pm arcCos\frac{\sqrt{2} }{2}+2\pi n,n\in Z\\\\8x=\pm \frac{\pi }{4}+2\pi n,n\in Z\\\\x=\pm \frac{\pi }{32}+\frac{\pi n }{4},n\ Z$

$1)x=\frac{\pi }{32}+\frac{\pi n }{4},n\in Z\\n=2\\\\x=\frac{\pi }{32}+\frac{2\pi }{4} =\frac{\pi }{32}+\frac{16\pi }{32}=\boxed{\frac{17\pi }{32}} \\\\n=3\\\\x=\frac{\pi }{32}+\frac{3\pi }{4}=\frac{\pi }{32}+\frac{24\pi }{32}=\boxed{\frac{25\pi }{32} }\\\\2)x=-\frac{\pi }{32}+\frac{\pi n }{4},n\in Z\\\\n=3\\\\x=-\frac{\pi }{32}+\frac{3\pi }{4}=-\frac{\pi }{32}+\frac{24\pi }{32}=\boxed{\frac{23\pi }{32}} \\\\n=4$

$x=-\frac{\pi }{32}+\frac{4\pi }{4}=-\frac{\pi }{32}+\frac{32\pi }{32}=\boxed{\frac{31\pi }{32}}$