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

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

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

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

x = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

8 u^{2} + 16 u + 8 = 0 : u = - 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 8, b = 16, c = 8

u_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 8 \cdot 8}{2 \cdot 8}

u_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 8 \cdot 8}{2 \cdot 8}

1 6^{2} - 4 \cdot 8 \cdot 8 = 0

1 6^{2} - 4 \cdot 8 \cdot 8

Multiply the numbers:

4 \cdot 8 \cdot 8 = 256

= 1 6^{2} - 256

1 6^{2} = 256

= 256 - 256

Subtract the numbers:

256 - 256 = 0

= 0

u_{1, 2} = \frac{- 16 \pm 0}{2 \cdot 8}

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

\frac{- 16}{2 \cdot 8} = - 1

\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

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = sin (x)

sin (x) = - 1

sin (x) = - 1

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{3 π}{2} + 2 πn

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ < 2 π

3 tan (x) = cot (x)

cos (θ) = \frac{7}{25}

2 tan^{2} (θ) sin (θ) - tan^{2} (θ) = 0

sin^{2} (x) = - 3 sin (x) + 4

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