Question

8) Прошу. Помогите!

a) sin 2x = $frac{1}{2}$
b) 1+2cosx=0

Answer 1

а) sin2x = 1/2
2x = (-1)^k * π/6 + πk, k∈Z
x = (-1)^k * π/12 + πk/2, k∈Z

б) 1 + 2cosx = 0
2cosx = -1
cosx = -1/2
x = +/- 2π/3 + 2πk, k∈Z