What is the general solution for 6sin^2(θ)+13sin(θ)+7=0 ?

The general solution for 6sin^2(θ)+13sin(θ)+7=0 is θ=(3pi)/2+2pin

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Popular >

6sin^2(θ)+13sin(θ)+7=0

Solution

6 sin^{2} (θ) + 13 sin (θ) + 7 = 0

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

6 sin^{2} (θ) + 13 sin (θ) + 7 = 0

Solve by substitution

6 sin^{2} (θ) + 13 sin (θ) + 7 = 0

Let:

sin (θ) = u

6 u^{2} + 13 u + 7 = 0

6 u^{2} + 13 u + 7 = 0 : u = - 1, u = - \frac{7}{6}

6 u^{2} + 13 u + 7 = 0

Solve with the quadratic formula

6 u^{2} + 13 u + 7 = 0

Quadratic Equation Formula:

For

a = 6, b = 13, c = 7

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 6 \cdot 7}{2 \cdot 6}

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 6 \cdot 7}{2 \cdot 6}

1 3^{2} - 4 \cdot 6 \cdot 7 = 1

1 3^{2} - 4 \cdot 6 \cdot 7

Multiply the numbers:

4 \cdot 6 \cdot 7 = 168

= 1 3^{2} - 168

1 3^{2} = 169

= 169 - 168

Subtract the numbers:

169 - 168 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 13 \pm 1}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 13 + 1}{2 \cdot 6}, u_{2} = \frac{- 13 - 1}{2 \cdot 6}

u = \frac{- 13 + 1}{2 \cdot 6} : - 1

\frac{- 13 + 1}{2 \cdot 6}

Add/Subtract the numbers:

- 13 + 1 = - 12

= \frac{- 12}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 12}{12}

Apply the fraction rule:

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

= - \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 13 - 1}{2 \cdot 6} : - \frac{7}{6}

\frac{- 13 - 1}{2 \cdot 6}

Subtract the numbers:

- 13 - 1 = - 14

= \frac{- 14}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 14}{12}

Apply the fraction rule:

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

= - \frac{14}{12}

Cancel the common factor:

2

= - \frac{7}{6}

The solutions to the quadratic equation are:

u = - 1, u = - \frac{7}{6}

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = - \frac{7}{6}

sin (θ) = - 1, sin (θ) = - \frac{7}{6}

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{7}{6} :

No Solution

sin (θ) = - \frac{7}{6}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

Popular Examples

12sin^2(x)-17sin(x)+6=0

12 sin^{2} (x) - 17 sin (x) + 6 = 0

sin(θ)=(sqrt(3))/3

sin (θ) = \frac{3}{3}

tan(x+pi)+cos(x+pi/2)=0

tan (x + π) + cos (x + \frac{π}{2}) = 0

sin(θ)cos(θ)=(sqrt(3))/4

sin (θ) cos (θ) = \frac{3}{4}

tan(x+pi/6)=sqrt(3)

tan (x + \frac{π}{6}) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 6sin^2(θ)+13sin(θ)+7=0 ?
The general solution for 6sin^2(θ)+13sin(θ)+7=0 is θ=(3pi)/2+2pin