Question

Решите уравнение

$2sin\frac{x}{2}+1=0$

Answer 1

$2sin \frac{x}{2}+1=0\\\\2sin \frac{x}{2}=-1\\\\sin \frac{x}{2}=- \frac{1}{2}\\\\ \frac{x}{2}=(-1)^{n+1}* \frac{ \pi }{6}+ \pi n, n\in Z\; \; |*2\\\\x=(-1)^{n+1}* \frac{ \pi }{3}+2 \pi n, n\in Z$

Answer 2

2sin$\frac{x}{2}$ + 1 = 0

2sin$\frac{x}{2}$  = -1

sin$\frac{x}{2}$  = -$\frac{1}{2}$

$\frac{x}{2}$ = (-1)^k arcsin(-$\frac{1}{2}$) + πk, k∈z

$\frac{x}{2}$ = (-1)^k+1arcsin$\frac{1}{2}$ + πk, k∈z

$\frac{x}{2}$ = (-1)^k+1$\frac{\pi}{6}$ + πk, k∈z

x = (-1)^k+1$\frac{\pi}{3}$ + 2πk, k∈z