What is the general solution for 12sin^2(x)+18sin(x)+6=0 ?

The general solution for 12sin^2(x)+18sin(x)+6=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin

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Popular >

12sin^2(x)+18sin(x)+6=0

Solution

12 sin^{2} (x) + 18 sin (x) + 6 = 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

12 sin^{2} (x) + 18 sin (x) + 6 = 0

Solve by substitution

12 sin^{2} (x) + 18 sin (x) + 6 = 0

Let:

sin (x) = u

12 u^{2} + 18 u + 6 = 0

12 u^{2} + 18 u + 6 = 0 : u = - \frac{1}{2}, u = - 1

12 u^{2} + 18 u + 6 = 0

Solve with the quadratic formula

12 u^{2} + 18 u + 6 = 0

Quadratic Equation Formula:

For

a = 12, b = 18, c = 6

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

u_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 12 \cdot 6}{2 \cdot 12}

1 8^{2} - 4 \cdot 12 \cdot 6 = 6

1 8^{2} - 4 \cdot 12 \cdot 6

Multiply the numbers:

4 \cdot 12 \cdot 6 = 288

= 1 8^{2} - 288

1 8^{2} = 324

= 324 - 288

Subtract the numbers:

324 - 288 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

u_{1, 2} = \frac{- 18 \pm 6}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- 18 + 6}{2 \cdot 12}, u_{2} = \frac{- 18 - 6}{2 \cdot 12}

u = \frac{- 18 + 6}{2 \cdot 12} : - \frac{1}{2}

\frac{- 18 + 6}{2 \cdot 12}

Add/Subtract the numbers:

- 18 + 6 = - 12

= \frac{- 12}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 12}{24}

Apply the fraction rule:

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

= - \frac{12}{24}

Cancel the common factor:

12

= - \frac{1}{2}

u = \frac{- 18 - 6}{2 \cdot 12} : - 1

\frac{- 18 - 6}{2 \cdot 12}

Subtract the numbers:

- 18 - 6 = - 24

= \frac{- 24}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 24}{24}

Apply the fraction rule:

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

= - \frac{24}{24}

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

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Frequently Asked Questions (FAQ)

What is the general solution for 12sin^2(x)+18sin(x)+6=0 ?
The general solution for 12sin^2(x)+18sin(x)+6=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x=(3pi)/2+2pin