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

The general solution for 2sin^2(u)=1-sin(u) is u= pi/6+2pin,u=(5pi)/6+2pin,u=(3pi)/2+2pin

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2sin^2(u)=1-sin(u)

Solution

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

Solution

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

Degrees

u = 3 0^{\circ} + 36 0^{\circ} n, u = 15 0^{\circ} + 36 0^{\circ} n, u = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (u) = v

2 v^{2} = 1 - v

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

2 v^{2} = 1 - v

Move

v

to the left side

2 v^{2} = 1 - v

Add

v

to both sides

2 v^{2} + v = 1 - v + v

Simplify

2 v^{2} + v = 1

2 v^{2} + v = 1

Move

1

to the left side

2 v^{2} + v = 1

Subtract

1

from both sides

2 v^{2} + v - 1 = 1 - 1

Simplify

2 v^{2} + v - 1 = 0

2 v^{2} + v - 1 = 0

Solve with the quadratic formula

2 v^{2} + v - 1 = 0

Quadratic Equation Formula:

For

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

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

v_{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

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 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

v_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

Separate the solutions

v_{1} = \frac{- 1 + 3}{2 \cdot 2}, v_{2} = \frac{- 1 - 3}{2 \cdot 2}

v = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

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

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

v = sin (u)

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

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

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

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

General solutions for

sin (u) = \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}

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

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

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

sin (u) = - 1

General solutions for

sin (u) = - 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}

u = \frac{3 π}{2} + 2 πn

u = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Popular Examples

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sin (t) = - \frac{3}{2}

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2 sin (2 θ) - 3 sin (θ) = 0

cos^2(x)+cos^2(x/2)-1/2 =0

cos^{2} (x) + cos^{2} (\frac{x}{2}) - \frac{1}{2} = 0

4cos^2(x)-1=0,0<= x<= 2pi

4 cos^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(u)=1-sin(u) ?
The general solution for 2sin^2(u)=1-sin(u) is u= pi/6+2pin,u=(5pi)/6+2pin,u=(3pi)/2+2pin