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

The general solution for n=(2tan^3(2θ)-1)^{1/2} is θ=(arctan(\sqrt[3]{(n^2+1)/2}))/(2)+(pik)/2

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n=(2tan^3(2θ)-1)^{1/2}

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

Solution steps

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

Switch sides

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

Square both sides:

2 tan^{3} (2 θ) - 1 = n^{2}

2 tan^{3} (2 θ) - 1 = n^{2}

Solve

Verify Solutions:

The solution is

Apply trig inverse properties

Solve

0 = 2 sin (x + 1) + 1

tan (θ) \cdot 10 = 10

4 cos (x) + tan (4 5^{\circ}) = 0.6

sin (5 x + 8) = cos (9 x - 16)

cos (x^{\circ}) = \frac{125}{174}

What is the general solution for n=(2tan^3(2θ)-1)^{1/2} ?
The general solution for n=(2tan^3(2θ)-1)^{1/2} is θ=(arctan(\sqrt[3]{(n^2+1)/2}))/(2)+(pik)/2