What is the general solution for 4sin^2(θ)-13sin(θ)+9=0 ?

The general solution for 4sin^2(θ)-13sin(θ)+9=0 is θ= pi/2+2pin

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4sin^2(θ)-13sin(θ)+9=0

Solution

4 sin^{2} (θ) - 13 sin (θ) + 9 = 0

Solution

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (θ) - 13 sin (θ) + 9 = 0

Solve by substitution

4 sin^{2} (θ) - 13 sin (θ) + 9 = 0

Let:

sin (θ) = u

4 u^{2} - 13 u + 9 = 0

4 u^{2} - 13 u + 9 = 0 : u = \frac{9}{4}, u = 1

4 u^{2} - 13 u + 9 = 0

Solve with the quadratic formula

4 u^{2} - 13 u + 9 = 0

Quadratic Equation Formula:

For

a = 4, b = - 13, c = 9

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

(- 13)^{2} - 4 \cdot 4 \cdot 9 = 5

(- 13)^{2} - 4 \cdot 4 \cdot 9

Apply exponent rule:

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

n

is even

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

= 1 3^{2} - 4 \cdot 4 \cdot 9

Multiply the numbers:

4 \cdot 4 \cdot 9 = 144

= 1 3^{2} - 144

1 3^{2} = 169

= 169 - 144

Subtract the numbers:

169 - 144 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 4}, u_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 4}

u = \frac{- ( - 13 ) + 5}{2 \cdot 4} : \frac{9}{4}

\frac{- ( - 13 ) + 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{13 + 5}{2 \cdot 4}

Add the numbers:

13 + 5 = 18

= \frac{18}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{18}{8}

Cancel the common factor:

2

= \frac{9}{4}

u = \frac{- ( - 13 ) - 5}{2 \cdot 4} : 1

\frac{- ( - 13 ) - 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{13 - 5}{2 \cdot 4}

Subtract the numbers:

13 - 5 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{9}{4}, u = 1

Substitute back

u = sin (θ)

sin (θ) = \frac{9}{4}, sin (θ) = 1

sin (θ) = \frac{9}{4}, sin (θ) = 1

sin (θ) = \frac{9}{4} :

No Solution

sin (θ) = \frac{9}{4}

- 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

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(θ)-13sin(θ)+9=0 ?
The general solution for 4sin^2(θ)-13sin(θ)+9=0 is θ= pi/2+2pin