What is the general solution for (1+2sin(2θ))^{1/3}=0 ?

The general solution for (1+2sin(2θ))^{1/3}=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin

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(1+2sin(2θ))^{1/3}=0

(1 + 2 sin (2 θ))^{\frac{1}{3}} = 0

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Solution steps

(1 + 2 sin (2 θ))^{\frac{1}{3}} = 0

Take both sides of the equation to the power of

3 : 1 + 2 sin (2 θ) = 0

1 + 2 sin (2 θ) = 0

Solve

1 + 2 sin (2 θ) = 0 : sin (2 θ) = - \frac{1}{2}

sin (2 θ) = - \frac{1}{2}

Verify Solutions:

sin (2 θ) = - \frac{1}{2}

True

The solution is

sin (2 θ) = - \frac{1}{2}

General solutions for

sin (2 θ) = - \frac{1}{2}

2 θ = \frac{7 π}{6} + 2 πn, 2 θ = \frac{11 π}{6} + 2 πn

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

Solve

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{12} + πn

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

sin (\frac{x}{3}) cos (6 x) - cos (\frac{x}{3}) sin (6 x) = 0

6 sin (3 x) = 5

x, 4 sin^{2} (x) = 5 - 4 cos (x)

3 (sin (x)) = (3) (cos (x))

sin (a) = 12500

What is the general solution for (1+2sin(2θ))^{1/3}=0 ?
The general solution for (1+2sin(2θ))^{1/3}=0 is θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin