What is the general solution for 2sin^2(x)+2sin(x)=0 ?

The general solution for 2sin^2(x)+2sin(x)=0 is x=2pin,x=pi+2pin,x=(3pi)/2+2pin

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2sin^2(x)+2sin(x)=0

Solution

2 sin^{2} (x) + 2 sin (x) = 0

Solution

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) + 2 sin (x) = 0

Solve by substitution

2 sin^{2} (x) + 2 sin (x) = 0

Let:

sin (x) = u

2 u^{2} + 2 u = 0

2 u^{2} + 2 u = 0 : u = 0, u = - 1

2 u^{2} + 2 u = 0

Solve with the quadratic formula

2 u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

2^{2} - 4 \cdot 2 \cdot 0 = 2

2^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

u = \frac{- 2 + 2}{2 \cdot 2} : 0

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

Add/Subtract the numbers:

- 2 + 2 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 2 - 2}{2 \cdot 2} : - 1

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

Subtract the numbers:

- 2 - 2 = - 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:

u = 0, u = - 1

Substitute back

u = sin (x)

sin (x) = 0, sin (x) = - 1

sin (x) = 0, sin (x) = - 1

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

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

sin (x) = - 1

General solutions for

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

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2 sin^{2} (x) = sin (x) + 1, 0 \leq x \leq 2 π

15tan(θ)-7=5tan(θ)-3

15 tan (θ) - 7 = 5 tan (θ) - 3

sin^2(x)-9cos(x)+9=0,0<= x<= 2pi

sin^{2} (x) - 9 cos (x) + 9 = 0, 0 \leq x \leq 2 π

-10cos(2x)-32cos(x)-22=0

- 10 cos (2 x) - 32 cos (x) - 22 = 0

solvefor θ,sin(θ)=-3/4 solve for

θ, sin (θ) = - \frac{3}{4}

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)+2sin(x)=0 ?
The general solution for 2sin^2(x)+2sin(x)=0 is x=2pin,x=pi+2pin,x=(3pi)/2+2pin