Sin(t-π/2)−cos(2π+t)=√3

Question

Sin(t-π/2)−cos(2π+t)=√3

Answer 1

Sin(t - π/2) - Cos(2π + t) = √3
- Sin(π/2 - t) - Cost = √3
- Cost - Cost = √3
- 2Cost = √3
Cost = - √3/2
t₁,₂ = + -arcCos(- √3/2) + 2πn, n ∈ z
t ₁,₂ = + - 5π/6 + 2πn, n ∈ z