What is the general solution for 2sin^2(θ)+9sin(θ)-5=0 ?

The general solution for 2sin^2(θ)+9sin(θ)-5=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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2sin^2(θ)+9sin(θ)-5=0

Solution

2 sin^{2} (θ) + 9 sin (θ) - 5 = 0

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (θ) + 9 sin (θ) - 5 = 0

Solve by substitution

2 sin^{2} (θ) + 9 sin (θ) - 5 = 0

Let:

sin (θ) = u

2 u^{2} + 9 u - 5 = 0

2 u^{2} + 9 u - 5 = 0 : u = \frac{1}{2}, u = - 5

2 u^{2} + 9 u - 5 = 0

Solve with the quadratic formula

2 u^{2} + 9 u - 5 = 0

Quadratic Equation Formula:

For

a = 2, b = 9, c = - 5

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

9^{2} - 4 \cdot 2 (- 5) = 11

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

Apply rule

- (- a) = a

= 9^{2} + 4 \cdot 2 \cdot 5

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 9^{2} + 40

9^{2} = 81

= 81 + 40

Add the numbers:

81 + 40 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 9 \pm 11}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 9 + 11}{2 \cdot 2}, u_{2} = \frac{- 9 - 11}{2 \cdot 2}

u = \frac{- 9 + 11}{2 \cdot 2} : \frac{1}{2}

\frac{- 9 + 11}{2 \cdot 2}

Add/Subtract the numbers:

- 9 + 11 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 9 - 11}{2 \cdot 2} : - 5

\frac{- 9 - 11}{2 \cdot 2}

Subtract the numbers:

- 9 - 11 = - 20

= \frac{- 20}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 20}{4}

Apply the fraction rule:

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

= - \frac{20}{4}

Divide the numbers:

\frac{20}{4} = 5

= - 5

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 5

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - 5

sin (θ) = \frac{1}{2}, sin (θ) = - 5

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

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}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - 5 :

No Solution

sin (θ) = - 5

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)+9sin(θ)-5=0 ?
The general solution for 2sin^2(θ)+9sin(θ)-5=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin