What is the general solution for 2sin^2(x/2)-1=sin(x/2) ?

The general solution for 2sin^2(x/2)-1=sin(x/2) is x=pi+4pin,x=(7pi)/3+4pin,x=(11pi)/3+4pin

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2sin^2(x/2)-1=sin(x/2)

Solution

2 sin^{2} (\frac{x}{2}) - 1 = sin (\frac{x}{2})

Solution

x = π + 4 πn, x = \frac{7 π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn

Degrees

x = 18 0^{\circ} + 72 0^{\circ} n, x = 42 0^{\circ} + 72 0^{\circ} n, x = 66 0^{\circ} + 72 0^{\circ} n

Solution steps

2 sin^{2} (\frac{x}{2}) - 1 = sin (\frac{x}{2})

Solve by substitution

2 sin^{2} (\frac{x}{2}) - 1 = sin (\frac{x}{2})

Let:

sin (\frac{x}{2}) = u

2 u^{2} - 1 = u

2 u^{2} - 1 = u : u = 1, u = - \frac{1}{2}

2 u^{2} - 1 = u

Move

u

to the left side

2 u^{2} - 1 = u

Subtract

u

from both sides

2 u^{2} - 1 - u = u - u

Simplify

2 u^{2} - 1 - u = 0

2 u^{2} - 1 - u = 0

Write in the standard form

a x^{2} + b x + c = 0

2 u^{2} - u - 1 = 0

Solve with the quadratic formula

2 u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = - 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 1) = 3

(- 1)^{2} - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 2}

u = \frac{- ( - 1 ) + 3}{2 \cdot 2} : 1

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

Apply rule

- (- a) = a

= \frac{1 + 3}{2 \cdot 2}

Add the numbers:

1 + 3 = 4

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

= \frac{1 - 3}{2 \cdot 2}

Subtract the numbers:

1 - 3 = - 2

= \frac{- 2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

The solutions to the quadratic equation are:

u = 1, u = - \frac{1}{2}

Substitute back

u = sin (\frac{x}{2})

sin (\frac{x}{2}) = 1, sin (\frac{x}{2}) = - \frac{1}{2}

sin (\frac{x}{2}) = 1, sin (\frac{x}{2}) = - \frac{1}{2}

sin (\frac{x}{2}) = 1 : x = π + 4 πn

sin (\frac{x}{2}) = 1

General solutions for

sin (\frac{x}{2}) = 1

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{x}{2} = \frac{π}{2} + 2 πn

\frac{x}{2} = \frac{π}{2} + 2 πn

Solve

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

\frac{x}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

x = π + 4 πn

x = π + 4 πn

x = π + 4 πn

x = π + 4 πn

sin (\frac{x}{2}) = - \frac{1}{2} : x = \frac{7 π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn

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

General solutions for

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

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

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

Solve

\frac{x}{2} = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{3} + 4 πn

\frac{x}{2} = \frac{7 π}{6} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{7 π}{6} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn : \frac{7 π}{3} + 4 πn

2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{6} = \frac{7 π}{3}

2 \cdot \frac{7 π}{6}

Multiply fractions:

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

= \frac{7 π 2}{6}

Multiply the numbers:

7 \cdot 2 = 14

= \frac{14 π}{6}

Cancel the common factor:

2

= \frac{7 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{3} + 4 πn

x = \frac{7 π}{3} + 4 πn

x = \frac{7 π}{3} + 4 πn

x = \frac{7 π}{3} + 4 πn

Solve

\frac{x}{2} = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{3} + 4 πn

\frac{x}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

\frac{2 x}{2} = 2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} = 2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn : \frac{11 π}{3} + 4 πn

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{11 π}{6} = \frac{11 π}{3}

2 \cdot \frac{11 π}{6}

Multiply fractions:

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

= \frac{11 π 2}{6}

Multiply the numbers:

11 \cdot 2 = 22

= \frac{22 π}{6}

Cancel the common factor:

2

= \frac{11 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{7 π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn

Combine all the solutions

x = π + 4 πn, x = \frac{7 π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x/2)-1=sin(x/2) ?
The general solution for 2sin^2(x/2)-1=sin(x/2) is x=pi+4pin,x=(7pi)/3+4pin,x=(11pi)/3+4pin