Question

Решите простейшее тригонометрическое уравнение..

Answer 1

A)  sin(π - x) + 2cos(3π/2 - x) = √2/2
sinx - 2sinx = √2/2
sinx = -√2/2
x = (-1)^(n)arcsin(-√2/2) + 2πn, n∈Z
x = (-1)^(n+1)arcsin(√2/2) + 2πn, n∈Z
x = (-1)^(n+1)(π/4)+ 2πn, n∈Z

b)  (sinx/2 - 1)*(ctgx + √3) = 0
1)  sinx/2 - 1 = 0
sinx/2 = 1
x/2 = π/2 + 2πn, n∈z
x₁ = π + 4πn, n∈Z
2)  ctgx + √3 = 0
ctgx = - √3
x = arcctg(- √3) + πk, k∈Z
x₂ = 5π/6 + πk, k∈Z