What is the general solution for 16sin^2(x)+24sin(x)+8=0 ?

The general solution for 16sin^2(x)+24sin(x)+8=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin

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Popular >

16sin^2(x)+24sin(x)+8=0

Solution

16 sin^{2} (x) + 24 sin (x) + 8 = 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

16 sin^{2} (x) + 24 sin (x) + 8 = 0

Solve by substitution

16 sin^{2} (x) + 24 sin (x) + 8 = 0

Let:

sin (x) = u

16 u^{2} + 24 u + 8 = 0

16 u^{2} + 24 u + 8 = 0 : u = - \frac{1}{2}, u = - 1

16 u^{2} + 24 u + 8 = 0

Solve with the quadratic formula

16 u^{2} + 24 u + 8 = 0

Quadratic Equation Formula:

For

a = 16, b = 24, c = 8

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 16 \cdot 8}{2 \cdot 16}

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 16 \cdot 8}{2 \cdot 16}

2 4^{2} - 4 \cdot 16 \cdot 8 = 8

2 4^{2} - 4 \cdot 16 \cdot 8

Multiply the numbers:

4 \cdot 16 \cdot 8 = 512

= 2 4^{2} - 512

2 4^{2} = 576

= 576 - 512

Subtract the numbers:

576 - 512 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 24 \pm 8}{2 \cdot 16}

Separate the solutions

u_{1} = \frac{- 24 + 8}{2 \cdot 16}, u_{2} = \frac{- 24 - 8}{2 \cdot 16}

u = \frac{- 24 + 8}{2 \cdot 16} : - \frac{1}{2}

\frac{- 24 + 8}{2 \cdot 16}

Add/Subtract the numbers:

- 24 + 8 = - 16

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

Multiply the numbers:

2 \cdot 16 = 32

= \frac{- 16}{32}

Apply the fraction rule:

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

= - \frac{16}{32}

Cancel the common factor:

16

= - \frac{1}{2}

u = \frac{- 24 - 8}{2 \cdot 16} : - 1

\frac{- 24 - 8}{2 \cdot 16}

Subtract the numbers:

- 24 - 8 = - 32

= \frac{- 32}{2 \cdot 16}

Multiply the numbers:

2 \cdot 16 = 32

= \frac{- 32}{32}

Apply the fraction rule:

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

= - \frac{32}{32}

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

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Frequently Asked Questions (FAQ)

What is the general solution for 16sin^2(x)+24sin(x)+8=0 ?
The general solution for 16sin^2(x)+24sin(x)+8=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin