What is the general solution for 8sin^2(θ)+17sin(θ)+9=0 ?

The general solution for 8sin^2(θ)+17sin(θ)+9=0 is θ=(3pi)/2+2pin

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Popular >

8sin^2(θ)+17sin(θ)+9=0

Solution

8 sin^{2} (θ) + 17 sin (θ) + 9 = 0

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

8 sin^{2} (θ) + 17 sin (θ) + 9 = 0

Solve by substitution

8 sin^{2} (θ) + 17 sin (θ) + 9 = 0

Let:

sin (θ) = u

8 u^{2} + 17 u + 9 = 0

8 u^{2} + 17 u + 9 = 0 : u = - 1, u = - \frac{9}{8}

8 u^{2} + 17 u + 9 = 0

Solve with the quadratic formula

8 u^{2} + 17 u + 9 = 0

Quadratic Equation Formula:

For

a = 8, b = 17, c = 9

u_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 8 \cdot 9}{2 \cdot 8}

u_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 8 \cdot 9}{2 \cdot 8}

1 7^{2} - 4 \cdot 8 \cdot 9 = 1

1 7^{2} - 4 \cdot 8 \cdot 9

Multiply the numbers:

4 \cdot 8 \cdot 9 = 288

= 1 7^{2} - 288

1 7^{2} = 289

= 289 - 288

Subtract the numbers:

289 - 288 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 17 \pm 1}{2 \cdot 8}

Separate the solutions

u_{1} = \frac{- 17 + 1}{2 \cdot 8}, u_{2} = \frac{- 17 - 1}{2 \cdot 8}

u = \frac{- 17 + 1}{2 \cdot 8} : - 1

\frac{- 17 + 1}{2 \cdot 8}

Add/Subtract the numbers:

- 17 + 1 = - 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

u = \frac{- 17 - 1}{2 \cdot 8} : - \frac{9}{8}

\frac{- 17 - 1}{2 \cdot 8}

Subtract the numbers:

- 17 - 1 = - 18

= \frac{- 18}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 18}{16}

Apply the fraction rule:

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

= - \frac{18}{16}

Cancel the common factor:

2

= - \frac{9}{8}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{9}{8}

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = - \frac{9}{8}

sin (θ) = - 1, sin (θ) = - \frac{9}{8}

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

sin (θ) = - 1

General solutions for

sin (θ) = - 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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

sin (θ) = - \frac{9}{8} :

No Solution

sin (θ) = - \frac{9}{8}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

Popular Examples

sin(2θ)=(sqrt(2))/2

sin (2 θ) = \frac{2}{2}

sec^2(x)-4tan(x)=-2

sec^{2} (x) - 4 tan (x) = - 2

-1=tan(θ)

- 1 = tan (θ)

4cos(x)=2

4 cos (x) = 2

5+2tan(3θ+pi/3)=3

5 + 2 tan (3 θ + \frac{π}{3}) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 8sin^2(θ)+17sin(θ)+9=0 ?
The general solution for 8sin^2(θ)+17sin(θ)+9=0 is θ=(3pi)/2+2pin