Question

укажите все корни уравнения sin(3π/2 - x)=1/2, на промежутке [-π;π]

Answer 1

$Sin\Big(\dfrac{3\pi }{2}-x\Big)=\dfrac{1}{2}\\\\-Cosx=\dfrac{1}{2}\\\\Cosx=-\dfrac{1}{2} \\\\x=\pm arcCos\Big(-\dfrac{1}{2}\Big)+2\pi n,n\in Z \\\\x=\pm \dfrac{2\pi }{3} +2\pi n,n\in Z \\\\1)x=-\dfrac{2\pi }{3} +2\pi n,n\in Z\\\\-\pi\leq-\dfrac{2\pi }{3}+2\pi n\leq \pi \\\\-\frac{\pi }{3} \leq 2\pi n\leq\dfrac{5\pi }{3}\\\\-\frac{1}{6}\leq n\leq \dfrac{5}{6} \\\\n=0 \ \Rightarrow \ -\dfrac{2\pi }{3}$

$2)x=\dfrac{2\pi }{3} +2\pi n,n\in Z\\\\-\pi\leq\dfrac{2\pi }{3}+2\pi n\leq \pi \\\\-\dfrac{5\pi }{3} \leq 2\pi n\leq\dfrac{\pi }{3}\\\\-\dfrac{5}{6}\leq n\leq \dfrac{1}{6}\\\\n=0 \ \Rightarrow \ x=\dfrac{2\pi }{3} \\\\Otvet:\boxed{-\dfrac{2\pi }{3} \ ; \ \dfrac{2\pi }{3}}$