Cos (2pi-2x)+3sin (pi-x)=2

Question

Cos (2pi-2x)+3sin (pi-x)=2

Answer 1

Cos(2П-2x)+3sin(П-x)=2
cos2x + 3sinx = 2
1 - 2sin²x + 3sinx - 2 = 0
2sin²x - 3sinx + 1 = 0
sinx = t
2t² - 3t + 1 = 0
D = 9 - 4*2*1 = 1
t₁ = (3 - 1)/4
t₁ = 1/2
t₂ = (3 + 1)/4
t₂ = 1
1)  sinx = 1/2
x = (-1)^n *arcsin(1/2) + πk, k∈Z
x₁ =  (-1)^n* (π/6) + πk, k∈Z
2)  sinx = 1
x₂ = π/2 + πn, n∈Z



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