What is the general solution for sin^2(x)+8sin(x)-9=0 ?

The general solution for sin^2(x)+8sin(x)-9=0 is x= pi/2+2pin

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Popular >

sin^2(x)+8sin(x)-9=0

Solution

sin^{2} (x) + 8 sin (x) - 9 = 0

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) + 8 sin (x) - 9 = 0

Solve by substitution

sin^{2} (x) + 8 sin (x) - 9 = 0

Let:

sin (x) = u

u^{2} + 8 u - 9 = 0

u^{2} + 8 u - 9 = 0 : u = 1, u = - 9

u^{2} + 8 u - 9 = 0

Solve with the quadratic formula

u^{2} + 8 u - 9 = 0

Quadratic Equation Formula:

For

a = 1, b = 8, c = - 9

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 9) = 10

8^{2} - 4 \cdot 1 \cdot (- 9)

Apply rule

- (- a) = a

= 8^{2} + 4 \cdot 1 \cdot 9

Multiply the numbers:

4 \cdot 1 \cdot 9 = 36

= 8^{2} + 36

8^{2} = 64

= 64 + 36

Add the numbers:

64 + 36 = 100

= 100

Factor the number:

100 = 1 0^{2}

= 1 0^{2}

Apply radical rule:

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 8 \pm 10}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 8 + 10}{2 \cdot 1}, u_{2} = \frac{- 8 - 10}{2 \cdot 1}

u = \frac{- 8 + 10}{2 \cdot 1} : 1

\frac{- 8 + 10}{2 \cdot 1}

Add/Subtract the numbers:

- 8 + 10 = 2

= \frac{2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- 8 - 10}{2 \cdot 1} : - 9

\frac{- 8 - 10}{2 \cdot 1}

Subtract the numbers:

- 8 - 10 = - 18

= \frac{- 18}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 18}{2}

Apply the fraction rule:

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

= - \frac{18}{2}

Divide the numbers:

\frac{18}{2} = 9

= - 9

The solutions to the quadratic equation are:

u = 1, u = - 9

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 9

sin (x) = 1, sin (x) = - 9

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = - 9 :

No Solution

sin (x) = - 9

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

Popular Examples

-2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ), 0^{\circ} \leq θ < 36 0^{\circ}

cot^2(θ)+4csc(θ)=-5

cot^{2} (θ) + 4 csc (θ) = - 5

10sin(a)+7sin(a)+1sin(a)+8sin(a)=48

10 sin (a) + 7 sin (a) + 1 sin (a) + 8 sin (a) = 48

2cos^2(θ)+7sin(θ)=5

2 cos^{2} (θ) + 7 sin (θ) = 5

(2sqrt(3))/3 =sec(θ)

\frac{2 3}{3} = sec (θ)

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)+8sin(x)-9=0 ?
The general solution for sin^2(x)+8sin(x)-9=0 is x= pi/2+2pin