What is the general solution for 7cos^2(θ)+6cos(θ)-12=7cos(θ)-6 ?

The general solution for 7cos^2(θ)+6cos(θ)-12=7cos(θ)-6 is θ=2pin,θ=2.60049…+2pin,θ=-2.60049…+2pin

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7cos^2(θ)+6cos(θ)-12=7cos(θ)-6

Solution

7 cos^{2} (θ) + 6 cos (θ) - 12 = 7 cos (θ) - 6

Solution

θ = 2 πn, θ = 2.60049 \dots + 2 πn, θ = - 2.60049 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 148.99728 \dots^{\circ} + 36 0^{\circ} n, θ = - 148.99728 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 cos^{2} (θ) + 6 cos (θ) - 12 = 7 cos (θ) - 6

Solve by substitution

7 cos^{2} (θ) + 6 cos (θ) - 12 = 7 cos (θ) - 6

Let:

cos (θ) = u

7 u^{2} + 6 u - 12 = 7 u - 6

7 u^{2} + 6 u - 12 = 7 u - 6 : u = 1, u = - \frac{6}{7}

7 u^{2} + 6 u - 12 = 7 u - 6

Move

6

to the left side

7 u^{2} + 6 u - 12 = 7 u - 6

Add

6

to both sides

7 u^{2} + 6 u - 12 + 6 = 7 u - 6 + 6

Simplify

7 u^{2} + 6 u - 6 = 7 u

7 u^{2} + 6 u - 6 = 7 u

Move

7 u

to the left side

7 u^{2} + 6 u - 6 = 7 u

Subtract

7 u

from both sides

7 u^{2} + 6 u - 6 - 7 u = 7 u - 7 u

Simplify

7 u^{2} - u - 6 = 0

7 u^{2} - u - 6 = 0

Solve with the quadratic formula

7 u^{2} - u - 6 = 0

Quadratic Equation Formula:

For

a = 7, b = - 1, c = - 6

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 7 ( - 6 )}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 7 ( - 6 )}{2 \cdot 7}

(- 1)^{2} - 4 \cdot 7 (- 6) = 13

(- 1)^{2} - 4 \cdot 7 (- 6)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 7 \cdot 6

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 7 \cdot 6 = 168

4 \cdot 7 \cdot 6

Multiply the numbers:

4 \cdot 7 \cdot 6 = 168

= 168

= 1 + 168

Add the numbers:

1 + 168 = 169

= 169

Factor the number:

169 = 1 3^{2}

= 1 3^{2}

Apply radical rule:

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 7}, u_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 7}

u = \frac{- ( - 1 ) + 13}{2 \cdot 7} : 1

\frac{- ( - 1 ) + 13}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{1 + 13}{2 \cdot 7}

Add the numbers:

1 + 13 = 14

= \frac{14}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{14}{14}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 1 ) - 13}{2 \cdot 7} : - \frac{6}{7}

\frac{- ( - 1 ) - 13}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{1 - 13}{2 \cdot 7}

Subtract the numbers:

1 - 13 = - 12

= \frac{- 12}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 12}{14}

Apply the fraction rule:

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

= - \frac{12}{14}

Cancel the common factor:

2

= - \frac{6}{7}

The solutions to the quadratic equation are:

u = 1, u = - \frac{6}{7}

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = - \frac{6}{7}

cos (θ) = 1, cos (θ) = - \frac{6}{7}

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

cos (θ) = - \frac{6}{7} : θ = arccos (- \frac{6}{7}) + 2 πn, θ = - arccos (- \frac{6}{7}) + 2 πn

cos (θ) = - \frac{6}{7}

Apply trig inverse properties

cos (θ) = - \frac{6}{7}

General solutions for

cos (θ) = - \frac{6}{7}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{6}{7}) + 2 πn, θ = - arccos (- \frac{6}{7}) + 2 πn

θ = arccos (- \frac{6}{7}) + 2 πn, θ = - arccos (- \frac{6}{7}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = arccos (- \frac{6}{7}) + 2 πn, θ = - arccos (- \frac{6}{7}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = 2.60049 \dots + 2 πn, θ = - 2.60049 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7cos^2(θ)+6cos(θ)-12=7cos(θ)-6 ?
The general solution for 7cos^2(θ)+6cos(θ)-12=7cos(θ)-6 is θ=2pin,θ=2.60049…+2pin,θ=-2.60049…+2pin