What is the general solution for 2sin(3θ)=1 ?

The general solution for 2sin(3θ)=1 is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3

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2sin(3θ)=1

Solution

2 sin (3 θ) = 1

Solution

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

Degrees

θ = 1 0^{\circ} + 12 0^{\circ} n, θ = 5 0^{\circ} + 12 0^{\circ} n

Solution steps

2 sin (3 θ) = 1

Divide both sides by

2

2 sin (3 θ) = 1

Divide both sides by

2

\frac{2 sin ( 3 θ )}{2} = \frac{1}{2}

Simplify

sin (3 θ) = \frac{1}{2}

sin (3 θ) = \frac{1}{2}

General solutions for

sin (3 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}

3 θ = \frac{π}{6} + 2 πn, 3 θ = \frac{5 π}{6} + 2 πn

3 θ = \frac{π}{6} + 2 πn, 3 θ = \frac{5 π}{6} + 2 πn

Solve

3 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{18} + \frac{2 πn}{3}

3 θ = \frac{π}{6} + 2 πn

Divide both sides by

3

3 θ = \frac{π}{6} + 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{π}{6}}{3} + \frac{2 πn}{3} : \frac{π}{18} + \frac{2 πn}{3}

\frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

Solve

3 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{18} + \frac{2 πn}{3}

3 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

3

3 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3} : \frac{5 π}{18} + \frac{2 πn}{3}

\frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{5 π}{6}}{3} = \frac{5 π}{18}

\frac{\frac{5 π}{6}}{3}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{5 π}{18}

= \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(3θ)=1 ?
The general solution for 2sin(3θ)=1 is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3