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= pi/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{π}{2} + 2 πn

Degrees

θ = 9 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 = \frac{7}{6}, u = 1

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 ( - 13 ) ^{2} - 4 \cdot 6 \cdot 7}{2 \cdot 6}

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

(- 13)^{2} - 4 \cdot 6 \cdot 7 = 1

(- 13)^{2} - 4 \cdot 6 \cdot 7

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 13)^{2} = 1 3^{2}

= 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} : \frac{7}{6}

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

Apply rule

- (- a) = a

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

Add the numbers:

13 + 1 = 14

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{14}{12}

Cancel the common factor:

2

= \frac{7}{6}

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

13 - 1 = 12

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = \frac{7}{6}, sin (θ) = 1

sin (θ) = \frac{7}{6}, sin (θ) = 1

sin (θ) = \frac{7}{6} :

No Solution

sin (θ) = \frac{7}{6}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = 1 : θ = \frac{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn

Popular Examples

sin(x)=0.1

sin (x) = 0.1

3sin^2(x)+6sin(x)+3=0

3 sin^{2} (x) + 6 sin (x) + 3 = 0

3sin(x)-4=sin(x)-2

3 sin (x) - 4 = sin (x) - 2

sin(θ)=-5/6

sin (θ) = - \frac{5}{6}

sec(θ)=0

sec (θ) = 0

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= pi/2+2pin