Question

sin2x+1=cosx+2sinx
СРОЧНО ДАМ 70 БАЛЛОВ​

Answer 1

$\sin2x+1=\cos x+2\sin x\\2\sin x\cos x+1=\cos x+2\sin x\\2\sin x\cos x-2\sin x-\cos x+1=0\\2\sin x(\cos x-1)-(\cos x-1)=0\\(\cos x-1)(2\sin x-1)=0$

$\left[\begin{array}{l} \cos x-1=0 \\ 2\sin x-1=0 \end{array}$

$\left[\begin{array}{l} \cos x=1 \\ \sin x=\dfrac{1}{2}\end{array}$

$\left[\begin{array}{l} x_1=2\pi n, \ n\in\mathbb{Z} \\ x_2=(-1)^k\dfrac{\pi}{6}+\pi k, \ k\in\mathbb{Z} \end{array}$

Ответ: $2\pi n$ и $(-1)^k\dfrac{\pi}{6}+\pi k$, где n и k - целые числа