f.4 a.maths

Solve the following equations for 0≦θ≦2π

Tan^ 2 θ= 2 + 4 cos ^ 2 θ

?? how to do?

tan2θ = 2 + 4cos2θ

(sinθ/cosθ)2 = 2 + 4cos2θ

sin2θ = 2cos2θ + 4cos4θ 【Assume cos2θ ≠ 0】

(1 - cos2θ) = 2cos2θ + 4cos4θ

4cos4θ + 3cos2θ - 1 = 0

4(cos2θ)2 + 3cos2θ - 1 = 0

(4cos2θ - 1)(cos2θ + 1) = 0

4cos2θ - 1 = 0 or cos2θ + 1 = 0

cos2θ = 1/4 or cos2θ = -1 (rejected as cos2θ > 0)

cosθ = ±1/2

For cosθ = 1/2

θ = π/3 or 5π/3

For cosθ = -1/2

θ = 2π/3 or θ = 4π/3

So the solution is θ = π/3, 2π/3, 4π/3 or 5π/3