Answer:
Solving for the unknown variable:
cosθ = sinθ
Dividing both sides by cosθ, we get:
tanθ = 1
Taking the inverse tangent of both sides, we get:
θ = π/4 + nπ, where n is an integer.
Therefore, the exact general solution is:
θ = π/4 + nπ, where n is an integer.
1 + sinθ = 2cos^2 (θ)
Using the identity cos^2 (θ) + sin^2 (θ) = 1, we can rewrite the equation as:
1 + sinθ = 2(1 - sin^2 (θ))
Expanding the right-hand side, we get:
1 + sinθ = 2 - 2sin^2 (θ)
Rearranging, we get:
2sin^2 (θ) + sinθ - 1 = 0
Using the quadratic formula, we get:
sinθ = (-1 ± √5)/4
Taking the inverse sine of both solutions, we get:
θ = arcsin [(-1 ± √5)/4] + 2nπ or π - arcsin [(-1 ± √5)/4] + 2nπ, where n is an integer.
Therefore, the exact general solution is:
θ = arcsin [(-1 ± √5)/4] + 2nπ or π - arcsin [(-1 ± √5)/4] + 2nπ, where n is an integer.
sin3θ = -1
Taking the inverse sine of both sides, we get:
3θ = (-1)^n (π/2) + 2nπ, where n is an integer.
Dividing both sides by 3, we get:
θ = (-1)^n (π/6) + (2nπ)/3, where n is an integer.
Therefore, the exact general solution is:
θ = (-1)^n (π/6) + (2nπ)/3, where n is an integer.