Question

РЕШИТЕ УРАВНЕНИЕ СРОЧНО ПОЖАЛУЙСТА cosx-√3sinx=√2 ОТКУПЛЮСЬ ЕСЛИ НАДО!

Answer 1

$Cosx-\sqrt{3}Sinx=\sqrt{2} |:2\\\\\frac{1}{2}Cosx-\frac{\sqrt{3} }{2}Sinx=\frac{\sqrt{2} }{2}\\\\Cos\frac{\pi }{3}Cosx-Sin\frac{\pi }{3}Sinx=\frac{\sqrt{2} }{2}\\\\Cos(\frac{\pi }{3}+x)=\frac{\sqrt{2} }{2}\\\\\frac{\pi }{3}+x=\pm arcCos\frac{\sqrt{2} }{2}+2\pi n,n\in z\\\\\frac{\pi }{3}+x=\pm \frac{\pi }{4}+2\pi n,n\in z\\\\x=\pm\frac{\pi }{4}-\frac{\pi }{3}+2\pi n,n\in z\\\\x_{1}=\frac{\pi }{4}-\frac{\pi }{3}+2\pi n=-\frac{\pi }{12}+2\pi n,n\in z$

$x_{2}=-\frac{\pi }{4}-\frac{\pi }{3}+2\pi n,n\in z\\\\x_{2}=-\frac{7\pi }{12}+2\pi n,n\in z$