What is the general solution for tan^2(θ)=3 ?

The general solution for tan^2(θ)=3 is θ= pi/3+pin,θ=(2pi)/3+pin

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tan^2(θ)=3

tan^{2} (θ) = 3

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn

Degrees

θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) = 3

Solve by substitution

tan^{2} (θ) = 3

Let:

tan (θ) = u

u^{2} = 3

u^{2} = 3 : u = 3, u = - 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 3, u = - 3

Substitute back

u = tan (θ)

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3 : θ = \frac{π}{3} + πn

tan (θ) = 3

General solutions for

tan (θ) = 3

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

tan (θ) = - 3 : θ = \frac{2 π}{3} + πn

tan (θ) = - 3

General solutions for

tan (θ) = - 3

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

Combine all the solutions

θ = \frac{π}{3} + πn, θ = \frac{2 π}{3} + πn

2 cos (2 x) = 1

2 cos (x) + 1 = 0

cos^{2} (θ) - sin^{2} (θ) = 1 + sin (θ)

cos^{2} (x) + cos (x) = 0

cos (2 x) sin (x) = 0

What is the general solution for tan^2(θ)=3 ?
The general solution for tan^2(θ)=3 is θ= pi/3+pin,θ=(2pi)/3+pin