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

The general solution for 2sin(2θ)=1 is θ= pi/(12)+pin,θ=(5pi)/(12)+pin

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2sin(2θ)=1

Solution

2 sin (2 θ) = 1

Solution

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

Degrees

θ = 1 5^{\circ} + 18 0^{\circ} n, θ = 7 5^{\circ} + 18 0^{\circ} n

Solution steps

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}

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

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

Solve

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

2 θ = \frac{π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

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

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

θ = \frac{π}{12} + πn

θ = \frac{π}{12} + πn

θ = \frac{π}{12} + πn

Solve

2 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{12} + πn

2 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 θ}{2} = \frac{\frac{5 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{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

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

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{π}{12} + πn, θ = \frac{5 π}{12} + πn

Popular Examples

sin(pi/8)

sin (\frac{π}{8})

csc((3pi)/4)

csc (\frac{3 π}{4})

cos(2)

cos (2)

prove sin^2(θ)(1+cot^2(θ))=1 prove

sin^{2} (θ) (1 + cot^{2} (θ)) = 1

5cos(6x)+6=9

5 cos (6 x) + 6 = 9

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2θ)=1 ?
The general solution for 2sin(2θ)=1 is θ= pi/(12)+pin,θ=(5pi)/(12)+pin