Question

Помогите решить уравнение cos⁡x-sin⁡x=cos⁡2x

Answer 1

Cosx - Sinx = Cos2x

Cosx - Sinx - (Cos²x - Sin²x) = 0

(Cosx - Sinx) - (Cosx - Sinx)(Cosx + Sinx) = 0

(Cosx - Sinx)(1 - Cosx - Sinx) = 0

1) Cosx - Sinx = 0

Разделим обе части на Cosx ≠ 0, получим :

1 - tgx = 0

tgx = 1

$x = \frac{\pi}{4}+\pi n,$n∈z

2) 1 - Cosx - Sinx = 0

Cosx + Sinx = 1

Разделим обе части на корень из двух, получим :

$\frac{1}{\sqrt{2}}Cosx+\frac{1}{\sqrt{2}}Sinx=\frac{1}{\sqrt{2}}}\\\\Cos\frac{\pi}{4}Cosx+Sin\frac{\pi}{4}Sinx=\frac{1}{\sqrt{2}} \\\\Cos(x-\frac{\pi}{4})=\frac{1}{\sqrt{2}}\\\\x-\frac{\pi}{4}=+-arcCos\frac{1}{\sqrt{2}} +2\pi n\\\\ x-\frac{\pi}{4}=+-\frac{\pi}{4}+2\pi n\\\\x_{1}=\frac{\pi}{2} +2\pi n\\\\x_{2}=2\pi n$