What is the general solution for 8sin^2(x)+12sin(x)+4=0 ?

The general solution for 8sin^2(x)+12sin(x)+4=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin

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8sin^2(x)+12sin(x)+4=0

Solution

8 sin^{2} (x) + 12 sin (x) + 4 = 0

Solution

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

8 sin^{2} (x) + 12 sin (x) + 4 = 0

Solve by substitution

8 sin^{2} (x) + 12 sin (x) + 4 = 0

Let:

sin (x) = u

8 u^{2} + 12 u + 4 = 0

8 u^{2} + 12 u + 4 = 0 : u = - \frac{1}{2}, u = - 1

8 u^{2} + 12 u + 4 = 0

Solve with the quadratic formula

8 u^{2} + 12 u + 4 = 0

Quadratic Equation Formula:

For

a = 8, b = 12, c = 4

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 8 \cdot 4}{2 \cdot 8}

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 8 \cdot 4}{2 \cdot 8}

1 2^{2} - 4 \cdot 8 \cdot 4 = 4

1 2^{2} - 4 \cdot 8 \cdot 4

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 1 2^{2} - 128

1 2^{2} = 144

= 144 - 128

Subtract the numbers:

144 - 128 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

n a^{n} = a

4^{2} = 4

= 4

u_{1, 2} = \frac{- 12 \pm 4}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 12 + 4}{2 \cdot 8}, u_{2} = \frac{- 12 - 4}{2 \cdot 8}

u = \frac{- 12 + 4}{2 \cdot 8} : - \frac{1}{2}

\frac{- 12 + 4}{2 \cdot 8}

Add/Subtract the numbers:

- 12 + 4 = - 8

= \frac{- 8}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 8}{16}

Apply the fraction rule:

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

= - \frac{8}{16}

Cancel the common factor:

8

= - \frac{1}{2}

u = \frac{- 12 - 4}{2 \cdot 8} : - 1

\frac{- 12 - 4}{2 \cdot 8}

Subtract the numbers:

- 12 - 4 = - 16

= \frac{- 16}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 16}{16}

Apply the fraction rule:

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

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

General solutions for

sin (x) = - \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}

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

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 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 = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 8sin^2(x)+12sin(x)+4=0 ?
The general solution for 8sin^2(x)+12sin(x)+4=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin