What is the general solution for 5sin^2(θ)+11sin(θ)+6=0 ?

The general solution for 5sin^2(θ)+11sin(θ)+6=0 is θ=(3pi)/2+2pin

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Popular >

5sin^2(θ)+11sin(θ)+6=0

Solution

5 sin^{2} (θ) + 11 sin (θ) + 6 = 0

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (θ) + 11 sin (θ) + 6 = 0

Solve by substitution

5 sin^{2} (θ) + 11 sin (θ) + 6 = 0

Let:

sin (θ) = u

5 u^{2} + 11 u + 6 = 0

5 u^{2} + 11 u + 6 = 0 : u = - 1, u = - \frac{6}{5}

5 u^{2} + 11 u + 6 = 0

Solve with the quadratic formula

5 u^{2} + 11 u + 6 = 0

Quadratic Equation Formula:

For

a = 5, b = 11, c = 6

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

1 1^{2} - 4 \cdot 5 \cdot 6 = 1

1 1^{2} - 4 \cdot 5 \cdot 6

Multiply the numbers:

4 \cdot 5 \cdot 6 = 120

= 1 1^{2} - 120

1 1^{2} = 121

= 121 - 120

Subtract the numbers:

121 - 120 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 11 \pm 1}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 11 + 1}{2 \cdot 5}, u_{2} = \frac{- 11 - 1}{2 \cdot 5}

u = \frac{- 11 + 1}{2 \cdot 5} : - 1

\frac{- 11 + 1}{2 \cdot 5}

Add/Subtract the numbers:

- 11 + 1 = - 10

= \frac{- 10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 10}{10}

Apply the fraction rule:

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

= - \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 11 - 1}{2 \cdot 5} : - \frac{6}{5}

\frac{- 11 - 1}{2 \cdot 5}

Subtract the numbers:

- 11 - 1 = - 12

= \frac{- 12}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 12}{10}

Apply the fraction rule:

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

= - \frac{12}{10}

Cancel the common factor:

2

= - \frac{6}{5}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{6}{5}

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = - \frac{6}{5}

sin (θ) = - 1, sin (θ) = - \frac{6}{5}

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{6}{5} :

No Solution

sin (θ) = - \frac{6}{5}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(θ)+11sin(θ)+6=0 ?
The general solution for 5sin^2(θ)+11sin(θ)+6=0 is θ=(3pi)/2+2pin