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

The general solution for 2sin(piθ)=1 is θ= 1/6+2n,θ= 5/6+2n

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2sin(piθ)=1

Solution

2 sin (π θ) = 1

Solution

θ = \frac{1}{6} + 2 n, θ = \frac{5}{6} + 2 n

Degrees

θ = 9.54929 \dots^{\circ} + 114.59155 \dots^{\circ} n, θ = 47.74648 \dots^{\circ} + 114.59155 \dots^{\circ} n

Solution steps

2 sin (π θ) = 1

Divide both sides by

2

2 sin (π θ) = 1

Divide both sides by

2

\frac{2 sin ( π θ )}{2} = \frac{1}{2}

Simplify

sin (π θ) = \frac{1}{2}

sin (π θ) = \frac{1}{2}

General solutions for

sin (π 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}

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

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

Solve

π θ = \frac{π}{6} + 2 πn : θ = \frac{1}{6} + 2 n

π θ = \frac{π}{6} + 2 πn

Divide both sides by

π

π θ = \frac{π}{6} + 2 πn

Divide both sides by

π

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

Simplify

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

Simplify

\frac{π θ}{π} : θ

\frac{π θ}{π}

Cancel the common factor:

π

= θ

Simplify

\frac{\frac{π}{6}}{π} + \frac{2 πn}{π} : \frac{1}{6} + 2 n

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

\frac{\frac{π}{6}}{π} = \frac{1}{6}

\frac{\frac{π}{6}}{π}

Apply the fraction rule:

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

= \frac{π}{6 π}

Cancel the common factor:

π

= \frac{1}{6}

\frac{2 πn}{π} = 2 n

\frac{2 πn}{π}

Cancel the common factor:

π

= 2 n

= \frac{1}{6} + 2 n

θ = \frac{1}{6} + 2 n

θ = \frac{1}{6} + 2 n

θ = \frac{1}{6} + 2 n

Solve

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

π θ = \frac{5 π}{6} + 2 πn

Divide both sides by

π

π θ = \frac{5 π}{6} + 2 πn

Divide both sides by

π

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

Simplify

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

Simplify

\frac{π θ}{π} : θ

\frac{π θ}{π}

Cancel the common factor:

π

= θ

Simplify

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

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

\frac{\frac{5 π}{6}}{π} = \frac{5}{6}

\frac{\frac{5 π}{6}}{π}

Apply the fraction rule:

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

= \frac{5 π}{6 π}

Cancel the common factor:

π

= \frac{5}{6}

\frac{2 πn}{π} = 2 n

\frac{2 πn}{π}

Cancel the common factor:

π

= 2 n

= \frac{5}{6} + 2 n

θ = \frac{5}{6} + 2 n

θ = \frac{5}{6} + 2 n

θ = \frac{5}{6} + 2 n

θ = \frac{1}{6} + 2 n, θ = \frac{5}{6} + 2 n

Popular Examples

2sqrt(3)sin(x/2)=3

23 sin (\frac{x}{2}) = 3

2cos(θ)=-1

2 cos (θ) = - 1

sin(θ)= 2/5

sin (θ) = \frac{2}{5}

cos(2x)-cos(x)=0,0<= x<= 2pi

cos (2 x) - cos (x) = 0, 0 \leq x \leq 2 π

cos(θ)=2

cos (θ) = 2

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(piθ)=1 ?
The general solution for 2sin(piθ)=1 is θ= 1/6+2n,θ= 5/6+2n