Trig Help ?

Question

please leave steps if you can but i can figure it out with answers on what i did wrong

Solve

tan2theta =-sqrt 3    0 to 360 degrees

Answer 1

The trigonometric equation is tan (2θ) = - √3.

tan (2θ) = tan(- π/3)

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

2θ = nπ - (π/3)

⇒ θ = [nπ - (π/3)]/2.

If n = 0, θ = [(0)π - (π/3)]/2 = - (π/3)/2 = - π/6,

If n = 1, θ = [(1)π - (π/3)]/2 = [π - π/3]/2 = (2π/3)/2 = 2π/6 = π/3,

If n = 2, θ = [(2)π - (π/3)]/2 = [2π - π/3]/2 = (5π/3)/2 = 5π/6,

If n = 3, θ = [(3)π - (π/3)]/2 = [3π - π/3]/2 = (8π/3)/2 = 8π/6 = 4π/3,

If n = 4, θ = [(4)π - (π/3)]/2 = [4π - π/3]/2 = (11π/3)/2 = 11π/6.

Therefore, the solutions of the given equation are θ = π/3, θ = 5π/6, θ = 4π/3 and θ = 11π/6 in the interval [0, 2π).