What is the general solution for 10sin^2(x)+11sin(x)-8=0 ?

The general solution for 10sin^2(x)+11sin(x)-8=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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10sin^2(x)+11sin(x)-8=0

Solution

10 sin^{2} (x) + 11 sin (x) - 8 = 0

Solution

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

Degrees

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

Solution steps

10 sin^{2} (x) + 11 sin (x) - 8 = 0

Solve by substitution

10 sin^{2} (x) + 11 sin (x) - 8 = 0

Let:

sin (x) = u

10 u^{2} + 11 u - 8 = 0

10 u^{2} + 11 u - 8 = 0 : u = \frac{1}{2}, u = - \frac{8}{5}

10 u^{2} + 11 u - 8 = 0

Solve with the quadratic formula

10 u^{2} + 11 u - 8 = 0

Quadratic Equation Formula:

For

a = 10, b = 11, c = - 8

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 10 ( - 8 )}{2 \cdot 10}

u_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 10 ( - 8 )}{2 \cdot 10}

1 1^{2} - 4 \cdot 10 (- 8) = 21

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

Apply rule

- (- a) = a

= 1 1^{2} + 4 \cdot 10 \cdot 8

Multiply the numbers:

4 \cdot 10 \cdot 8 = 320

= 1 1^{2} + 320

1 1^{2} = 121

= 121 + 320

Add the numbers:

121 + 320 = 441

= 441

Factor the number:

441 = 2 1^{2}

= 2 1^{2}

Apply radical rule:

2 1^{2} = 21

= 21

u_{1, 2} = \frac{- 11 \pm 21}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- 11 + 21}{2 \cdot 10}, u_{2} = \frac{- 11 - 21}{2 \cdot 10}

u = \frac{- 11 + 21}{2 \cdot 10} : \frac{1}{2}

\frac{- 11 + 21}{2 \cdot 10}

Add/Subtract the numbers:

- 11 + 21 = 10

= \frac{10}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{10}{20}

Cancel the common factor:

10

= \frac{1}{2}

u = \frac{- 11 - 21}{2 \cdot 10} : - \frac{8}{5}

\frac{- 11 - 21}{2 \cdot 10}

Subtract the numbers:

- 11 - 21 = - 32

= \frac{- 32}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 32}{20}

Apply the fraction rule:

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

= - \frac{32}{20}

Cancel the common factor:

4

= - \frac{8}{5}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - \frac{8}{5}

Substitute back

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = - \frac{8}{5}

sin (x) = \frac{1}{2}, sin (x) = - \frac{8}{5}

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

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \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}

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

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

sin (x) = - \frac{8}{5} :

No Solution

sin (x) = - \frac{8}{5}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

Popular Examples

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2 cos^{2} (x) + 7 cos (x) - 4 = 0

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cot (2 θ) = \frac{3}{3}

4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

-cos(θ)+1=2cos(θ)+3

- cos (θ) + 1 = 2 cos (θ) + 3

cot^2(x)+cot(x)=0

cot^{2} (x) + cot (x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 10sin^2(x)+11sin(x)-8=0 ?
The general solution for 10sin^2(x)+11sin(x)-8=0 is x= pi/6+2pin,x=(5pi)/6+2pin