Question

find all real values of θ in the interval [0°,360°) that satisfy the equation

4sin^4θ - 5sin^2θ+1=0

Answer 1

Given equation is 4sin^4θ - 5sin^2θ + 1 = 0

Let t = sin^2θ

      t^2 = sin^4θ

Therefore the given equation can be written as 4t^2 - 5t +1 = 0

                                                                      => 4t^2 - 4t - t + 1 = 0

                                                                      => 4t (t - 1) -1(t - 1) = 0

                                                                      => (t - 1) (4t - 1) = 0

                                                                      => t =1 and t = 1/4

                                                                      => sin^2θ = 1 and sin^2θ = 1/4    [ Where t = sin^2θ ]

                                                                      => sinθ = √1 and sinθ = √(1/4)

                                                                      => sinθ = +/- 1 and sinθ = +/-1/2 

                                                                      => θ = π/2, 9π/6 and θ = π/6, 5π/6, 7π/6, 11π/6

The real values of θ in the interval [0, 360) are π/6, π/2, 5π/6, 7π/6, 9π/6, 11π/6.

 

Answer 2

The solution set is