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

Solution

(1 + 2 sin (2 θ))^{\frac{1}{3}} = 0

Solution

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))^{\frac{1}{3}} = 0

((1 + 2 sin (2 θ))^{\frac{1}{3}})^{3} = 0^{3}

Expand

((1 + 2 sin (2 θ))^{\frac{1}{3}})^{3} : 1 + 2 sin (2 θ)

((1 + 2 sin (2 θ))^{\frac{1}{3}})^{3}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

\frac{1}{3} \cdot 3 = 1

\frac{1}{3} \cdot 3

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 3}{3}

Cancel the common factor:

3

= 1

= 1 + 2 sin (2 θ)

Refine

= 1 + 2 sin (2 θ)

Expand

0^{3} : 0

0^{3}

Apply rule

0^{a} = 0

= 0

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}

1 + 2 sin (2 θ) = 0

Move

1

to the right side

1 + 2 sin (2 θ) = 0

Subtract

1

from both sides

1 + 2 sin (2 θ) - 1 = 0 - 1

Simplify

2 sin (2 θ) = - 1

2 sin (2 θ) = - 1

Divide both sides by

2

2 sin (2 θ) = - 1

Divide both sides by

2

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

Simplify

sin (2 θ) = - \frac{1}{2}

sin (2 θ) = - \frac{1}{2}

sin (2 θ) = - \frac{1}{2}

Verify Solutions:

sin (2 θ) = - \frac{1}{2}

True

Check the solutions by plugging them into

(1 + 2 sin (2 θ))^{\frac{1}{3}} = 0

Remove the ones that don't agree with the equation.

Plug in

sin (2 θ) = - \frac{1}{2} :

True

(1 + 2 (- \frac{1}{2}))^{\frac{1}{3}} = 0

(1 + 2 (- \frac{1}{2}))^{\frac{1}{3}} = 0

(1 + 2 (- \frac{1}{2}))^{\frac{1}{3}}

Remove parentheses:

(- a) = - a

= (1 - 2 \cdot \frac{1}{2})^{\frac{1}{3}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= (1 - 1)^{\frac{1}{3}}

Subtract the numbers:

1 - 1 = 0

= 0^{\frac{1}{3}}

Apply rule

0^{a} = 0

= 0

0 = 0

True

The solution is

sin (2 θ) = - \frac{1}{2}

General solutions for

sin (2 θ) = - \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{7 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

Solve

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

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

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

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Frequently Asked Questions (FAQ)

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