What is the general solution for sin(θ/2)=sqrt(2)-sin(θ/2) ?

The general solution for sin(θ/2)=sqrt(2)-sin(θ/2) is θ= pi/2+4pin,θ=(3pi)/2+4pin

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sin(θ/2)=sqrt(2)-sin(θ/2)

Solution

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

Solution

θ = \frac{π}{2} + 4 πn, θ = \frac{3 π}{2} + 4 πn

Degrees

θ = 9 0^{\circ} + 72 0^{\circ} n, θ = 27 0^{\circ} + 72 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (\frac{θ}{2}) = u

u = 2 - u

u = 2 - u : u = \frac{2}{2}

u = 2 - u

Move

u

to the left side

u = 2 - u

Add

u

to both sides

u + u = 2 - u + u

Simplify

2 u = 2

2 u = 2

Divide both sides by

2

2 u = 2

Divide both sides by

2

\frac{2 u}{2} = \frac{2}{2}

Simplify

u = \frac{2}{2}

u = \frac{2}{2}

Substitute back

u = sin (\frac{θ}{2})

sin (\frac{θ}{2}) = \frac{2}{2}

sin (\frac{θ}{2}) = \frac{2}{2}

sin (\frac{θ}{2}) = \frac{2}{2} : θ = \frac{π}{2} + 4 πn, θ = \frac{3 π}{2} + 4 πn

sin (\frac{θ}{2}) = \frac{2}{2}

General solutions for

sin (\frac{θ}{2}) = \frac{2}{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}

\frac{θ}{2} = \frac{π}{4} + 2 πn, \frac{θ}{2} = \frac{3 π}{4} + 2 πn

\frac{θ}{2} = \frac{π}{4} + 2 πn, \frac{θ}{2} = \frac{3 π}{4} + 2 πn

Solve

\frac{θ}{2} = \frac{π}{4} + 2 πn : θ = \frac{π}{2} + 4 πn

\frac{θ}{2} = \frac{π}{4} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{4} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{π}{4} + 2 \cdot 2 πn : \frac{π}{2} + 4 πn

2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

Multiply fractions:

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

= \frac{π 2}{4}

Cancel the common factor:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{2} + 4 πn

θ = \frac{π}{2} + 4 πn

θ = \frac{π}{2} + 4 πn

θ = \frac{π}{2} + 4 πn

Solve

\frac{θ}{2} = \frac{3 π}{4} + 2 πn : θ = \frac{3 π}{2} + 4 πn

\frac{θ}{2} = \frac{3 π}{4} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{3 π}{4} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn : \frac{3 π}{2} + 4 πn

2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{4} = \frac{3 π}{2}

2 \cdot \frac{3 π}{4}

Multiply fractions:

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

= \frac{3 π 2}{4}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{6 π}{4}

Cancel the common factor:

2

= \frac{3 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{3 π}{2} + 4 πn

θ = \frac{3 π}{2} + 4 πn

θ = \frac{3 π}{2} + 4 πn

θ = \frac{3 π}{2} + 4 πn

θ = \frac{π}{2} + 4 πn, θ = \frac{3 π}{2} + 4 πn

Combine all the solutions

θ = \frac{π}{2} + 4 πn, θ = \frac{3 π}{2} + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ/2)=sqrt(2)-sin(θ/2) ?
The general solution for sin(θ/2)=sqrt(2)-sin(θ/2) is θ= pi/2+4pin,θ=(3pi)/2+4pin