Question

Решить уравнение: 1+2sin(x)=0

Answer 1

Решение:
1 + 2sinx = 0
2sinx = - 1
sinx = -1/2
x = (- 1)^n• arcsin (- 1/2) + πn, n∊Z
x = (- 1)^n•(- π/6) + πn, n∊Z
x = (- 1)^(n+1) • π/6 + πn, n∊Z
Ответ: (- 1)^(n+1) • π/6 + πn, n∊Z.

Answer 2

$1 + 2sin x = 0 \ \ sin x =- dfrac{1}{2} \ \ left[ begin{gathered} x = -dfrac {pi}{6} + 2pi n, n in Z \ x = -dfrac{5pi}{6} + 2pi n, nin Z end{gathered} right.$

Ответ: x = -π/6 + 2πn, n ∈ Z ; x = -5π/6 + 2πn, n ∈ Z