What is the general solution for 8cos^2(x)-12sin(x)-12=0 ?

The general solution for 8cos^2(x)-12sin(x)-12=0 is x=(3pi)/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin

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8cos^2(x)-12sin(x)-12=0

Solution

8 cos^{2} (x) - 12 sin (x) - 12 = 0

Solution

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

Degrees

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

Solution steps

8 cos^{2} (x) - 12 sin (x) - 12 = 0

Rewrite using trig identities

- 12 - 12 sin (x) + 8 cos^{2} (x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 12 - 12 sin (x) + 8 (1 - sin^{2} (x))

Simplify

- 12 - 12 sin (x) + 8 (1 - sin^{2} (x)) : - 8 sin^{2} (x) - 12 sin (x) - 4

- 12 - 12 sin (x) + 8 (1 - sin^{2} (x))

Expand

8 (1 - sin^{2} (x)) : 8 - 8 sin^{2} (x)

8 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 1, c = sin^{2} (x)

= 8 \cdot 1 - 8 sin^{2} (x)

Multiply the numbers:

8 \cdot 1 = 8

= 8 - 8 sin^{2} (x)

= - 12 - 12 sin (x) + 8 - 8 sin^{2} (x)

Simplify

- 12 - 12 sin (x) + 8 - 8 sin^{2} (x) : - 8 sin^{2} (x) - 12 sin (x) - 4

- 12 - 12 sin (x) + 8 - 8 sin^{2} (x)

Group like terms

= - 12 sin (x) - 8 sin^{2} (x) - 12 + 8

Add/Subtract the numbers:

- 12 + 8 = - 4

= - 8 sin^{2} (x) - 12 sin (x) - 4

= - 8 sin^{2} (x) - 12 sin (x) - 4

= - 8 sin^{2} (x) - 12 sin (x) - 4

- 4 - 12 sin (x) - 8 sin^{2} (x) = 0

Solve by substitution

- 4 - 12 sin (x) - 8 sin^{2} (x) = 0

Let:

sin (x) = u

- 4 - 12 u - 8 u^{2} = 0

- 4 - 12 u - 8 u^{2} = 0 : u = - 1, u = - \frac{1}{2}

- 4 - 12 u - 8 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 8 u^{2} - 12 u - 4 = 0

Solve with the quadratic formula

- 8 u^{2} - 12 u - 4 = 0

Quadratic Equation Formula:

For

a = - 8, b = - 12, c = - 4

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 8 ) ( - 4 )}{2 ( - 8 )}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 8 ) ( - 4 )}{2 ( - 8 )}

(- 12)^{2} - 4 (- 8) (- 4) = 4

(- 12)^{2} - 4 (- 8) (- 4)

Apply rule

- (- a) = a

= (- 12)^{2} - 4 \cdot 8 \cdot 4

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 12)^{2} = 1 2^{2}

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

Multiply the numbers:

4 \cdot 8 \cdot 4 = 128

= 1 2^{2} - 128

1 2^{2} = 144

= 144 - 128

Subtract the numbers:

144 - 128 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 12 ) \pm 4}{2 ( - 8 )}

Separate the solutions

u_{1} = \frac{- ( - 12 ) + 4}{2 ( - 8 )}, u_{2} = \frac{- ( - 12 ) - 4}{2 ( - 8 )}

u = \frac{- ( - 12 ) + 4}{2 ( - 8 )} : - 1

\frac{- ( - 12 ) + 4}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{12 + 4}{- 2 \cdot 8}

Add the numbers:

12 + 4 = 16

= \frac{16}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{16}{- 16}

Apply the fraction rule:

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

= - \frac{16}{16}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 12 ) - 4}{2 ( - 8 )} : - \frac{1}{2}

\frac{- ( - 12 ) - 4}{2 ( - 8 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{12 - 4}{- 2 \cdot 8}

Subtract the numbers:

12 - 4 = 8

= \frac{8}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{8}{- 16}

Apply the fraction rule:

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

= - \frac{8}{16}

Cancel the common factor:

8

= - \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 8cos^2(x)-12sin(x)-12=0 ?
The general solution for 8cos^2(x)-12sin(x)-12=0 is x=(3pi)/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin