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

The general solution for cot^2(θ)= 1/3 is θ=1.04719…+pin,θ=2.09439…+pin

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cot^2(θ)= 1/3

cot^{2} (θ) = \frac{1}{3}

θ = 1.04719 \dots + πn, θ = 2.09439 \dots + πn

Degrees

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

Solution steps

cot^{2} (θ) = \frac{1}{3}

Solve by substitution

cot^{2} (θ) = \frac{1}{3}

Let:

cot (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 = cot (θ)

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

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

cot (θ) = \frac{1}{3} : θ = arccot (\frac{1}{3}) + πn

cot (θ) = \frac{1}{3}

Apply trig inverse properties

cot (θ) = \frac{1}{3}

General solutions for

cot (θ) = \frac{1}{3}

cot (x) = a \Rightarrow x = arccot (a) + πn

θ = arccot (\frac{1}{3}) + πn

θ = arccot (\frac{1}{3}) + πn

cot (θ) = - \frac{1}{3} : θ = arccot (- \frac{1}{3}) + πn

cot (θ) = - \frac{1}{3}

Apply trig inverse properties

cot (θ) = - \frac{1}{3}

General solutions for

cot (θ) = - \frac{1}{3}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

θ = arccot (- \frac{1}{3}) + πn

θ = arccot (- \frac{1}{3}) + πn

Combine all the solutions

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

Show solutions in decimal form

θ = 1.04719 \dots + πn, θ = 2.09439 \dots + πn

x, y^{3} - sin^{2} (x) = 1

- cos (2 x) - 4 sin^{2} (x) + 2 = 0

sec (θ) - 2 cos (θ) = 0

sin (θ) = \frac{1}{2}, 0 \leq θ \leq 4 π

8 sin (θ) cos (θ) tan (θ) = csc (θ)

What is the general solution for cot^2(θ)= 1/3 ?
The general solution for cot^2(θ)= 1/3 is θ=1.04719…+pin,θ=2.09439…+pin