What is the general solution for 10sin^2(x)+15sin(x)+5=0 ?

The general solution for 10sin^2(x)+15sin(x)+5=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin

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Popular >

10sin^2(x)+15sin(x)+5=0

Solution

10 sin^{2} (x) + 15 sin (x) + 5 = 0

Solution

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

10 sin^{2} (x) + 15 sin (x) + 5 = 0

Solve by substitution

10 sin^{2} (x) + 15 sin (x) + 5 = 0

Let:

sin (x) = u

10 u^{2} + 15 u + 5 = 0

10 u^{2} + 15 u + 5 = 0 : u = - \frac{1}{2}, u = - 1

10 u^{2} + 15 u + 5 = 0

Solve with the quadratic formula

10 u^{2} + 15 u + 5 = 0

Quadratic Equation Formula:

For

a = 10, b = 15, c = 5

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

1 5^{2} - 4 \cdot 10 \cdot 5 = 5

1 5^{2} - 4 \cdot 10 \cdot 5

Multiply the numbers:

4 \cdot 10 \cdot 5 = 200

= 1 5^{2} - 200

1 5^{2} = 225

= 225 - 200

Subtract the numbers:

225 - 200 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 15 \pm 5}{2 \cdot 10}

Separate the solutions

u_{1} = \frac{- 15 + 5}{2 \cdot 10}, u_{2} = \frac{- 15 - 5}{2 \cdot 10}

u = \frac{- 15 + 5}{2 \cdot 10} : - \frac{1}{2}

\frac{- 15 + 5}{2 \cdot 10}

Add/Subtract the numbers:

- 15 + 5 = - 10

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

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 10}{20}

Apply the fraction rule:

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

= - \frac{10}{20}

Cancel the common factor:

10

= - \frac{1}{2}

u = \frac{- 15 - 5}{2 \cdot 10} : - 1

\frac{- 15 - 5}{2 \cdot 10}

Subtract the numbers:

- 15 - 5 = - 20

= \frac{- 20}{2 \cdot 10}

Multiply the numbers:

2 \cdot 10 = 20

= \frac{- 20}{20}

Apply the fraction rule:

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

= - \frac{20}{20}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = - 1 : x = \frac{3 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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 10sin^2(x)+15sin(x)+5=0 ?
The general solution for 10sin^2(x)+15sin(x)+5=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin