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

The general solution for 3tan^2(θ)+1= 2/(tan^2(θ)) is θ=0.68471…+pin,θ=-0.68471…+pin

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3tan^2(θ)+1= 2/(tan^2(θ))

3 tan^{2} (θ) + 1 = \frac{2}{tan ^{2} ( θ )}

θ = 0.68471 \dots + πn, θ = - 0.68471 \dots + πn

Solution steps

3 tan^{2} (θ) + 1 = \frac{2}{tan ^{2} ( θ )}

Solve by substitution

tan (θ) = \frac{2}{3}, tan (θ) = - \frac{2}{3}, tan (θ) = i, tan (θ) = - i

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

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

tan (θ) = i :

No Solution

tan (θ) = - i :

No Solution

Combine all the solutions

θ = arctan (\frac{2}{3}) + πn, θ = arctan (- \frac{2}{3}) + πn

Show solutions in decimal form

θ = 0.68471 \dots + πn, θ = - 0.68471 \dots + πn

4 tan (x) = 2 sec^{2} (x)

sin (x) + cos (x) + cot (x) = csc (x)

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

cos (x) = \frac{3}{13}

3 tan^{3} (x) - 9 tan (x) = 0

What is the general solution for 3tan^2(θ)+1= 2/(tan^2(θ)) ?
The general solution for 3tan^2(θ)+1= 2/(tan^2(θ)) is θ=0.68471…+pin,θ=-0.68471…+pin