What is the general solution for tan^2(θ)= 1/3 ?

The general solution for tan^2(θ)= 1/3 is θ=0.52359…+pin,θ=-0.52359…+pin

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tan^2(θ)= 1/3

tan^{2} (θ) = \frac{1}{3}

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = - 3 0^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (θ) = \frac{1}{3}

Solve by substitution

tan^{2} (θ) = \frac{1}{3}

Let:

tan (θ) = u

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3} : u = \frac{1}{3}, u = - \frac{1}{3}

u^{2} = \frac{1}{3}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = tan (θ)

tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = \frac{1}{3} : θ = arctan (\frac{1}{3}) + πn

tan (θ) = \frac{1}{3}

Apply trig inverse properties

tan (θ) = \frac{1}{3}

General solutions for

tan (θ) = \frac{1}{3}

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{1}{3}) + πn

θ = arctan (\frac{1}{3}) + πn

tan (θ) = - \frac{1}{3} : θ = arctan (- \frac{1}{3}) + πn

tan (θ) = - \frac{1}{3}

Apply trig inverse properties

tan (θ) = - \frac{1}{3}

General solutions for

tan (θ) = - \frac{1}{3}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- \frac{1}{3}) + πn

θ = arctan (- \frac{1}{3}) + πn

Combine all the solutions

θ = arctan (\frac{1}{3}) + πn, θ = arctan (- \frac{1}{3}) + πn

Show solutions in decimal form

θ = 0.52359 \dots + πn, θ = - 0.52359 \dots + πn

3 cot (x) - 1 = 0

cos (x) + 1 = - cos (x)

sin (2 θ) = - 1

sin^{2} (x) = \frac{3}{4}

2 sin (θ) = 1

What is the general solution for tan^2(θ)= 1/3 ?
The general solution for tan^2(θ)= 1/3 is θ=0.52359…+pin,θ=-0.52359…+pin