Вопрос: [latex]cos(2x+ \pi /4)cos x - sin(2x+ \pi /4)sin x =- \sqrt{2}/2 [/latex]

Вопрос:

[latex]cos(2x+ \pi /4)cos x - sin(2x+ \pi /4)sin x =- \sqrt{2}/2 [/latex]

Ответы:

[latex]cos(2x+\frac{\pi}{4})cosx-sin(2x+\frac{\pi}{4})sinx=-\frac{\sqrt{2}}{2}[/latex] по формуле косинуса суммы [latex]cos(a+b)=cos cos b - sin a sin b[/latex] [latex]cos(2x+\frac{\pi}{4}+x)=-\frac{\sqrt{2}}{2}[/latex] [latex]2x+\frac{\pi}{4}+x=arccos(-\frac{\sqrt{2}}{2})+2\pi*n[/latex] [latex]3x+\frac{\pi}{4}=\pi-arccos\frac{\sqrt{2}}{2}+2\pi*n[/latex] [latex]3x+\frac{\pi}{4}=^+_-\frac{3\pi}{4}+2*\pi*n;[/latex] [latex]3x=-\frac{\pi}{4}^+_-\frac{3\pi}{4}+2*\pi*n;[/latex] [latex]x=-\frac{\pi}{12}^+_-\frac{3\pi}{12}+\frac{2\pi*n}{3}[/latex] n є Z