What is the general solution for 7cos^2(θ)+11cos(θ)=cos(θ)-3 ?

The general solution for 7cos^2(θ)+11cos(θ)=cos(θ)-3 is θ=2.01370…+2pin,θ=-2.01370…+2pin,θ=pi+2pin

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7cos^2(θ)+11cos(θ)=cos(θ)-3

Solution

7 cos^{2} (θ) + 11 cos (θ) = cos (θ) - 3

Solution

θ = 2.01370 \dots + 2 πn, θ = - 2.01370 \dots + 2 πn, θ = π + 2 πn

Degrees

θ = 115.37693 \dots^{\circ} + 36 0^{\circ} n, θ = - 115.37693 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

7 cos^{2} (θ) + 11 cos (θ) = cos (θ) - 3

Solve by substitution

7 cos^{2} (θ) + 11 cos (θ) = cos (θ) - 3

Let:

cos (θ) = u

7 u^{2} + 11 u = u - 3

7 u^{2} + 11 u = u - 3 : u = - \frac{3}{7}, u = - 1

7 u^{2} + 11 u = u - 3

Move

3

to the left side

7 u^{2} + 11 u = u - 3

Add

3

to both sides

7 u^{2} + 11 u + 3 = u - 3 + 3

Simplify

7 u^{2} + 11 u + 3 = u

7 u^{2} + 11 u + 3 = u

Move

u

to the left side

7 u^{2} + 11 u + 3 = u

Subtract

u

from both sides

7 u^{2} + 11 u + 3 - u = u - u

Simplify

7 u^{2} + 10 u + 3 = 0

7 u^{2} + 10 u + 3 = 0

Solve with the quadratic formula

7 u^{2} + 10 u + 3 = 0

Quadratic Equation Formula:

For

a = 7, b = 10, c = 3

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 7 \cdot 3}{2 \cdot 7}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 7 \cdot 3}{2 \cdot 7}

1 0^{2} - 4 \cdot 7 \cdot 3 = 4

1 0^{2} - 4 \cdot 7 \cdot 3

Multiply the numbers:

4 \cdot 7 \cdot 3 = 84

= 1 0^{2} - 84

1 0^{2} = 100

= 100 - 84

Subtract the numbers:

100 - 84 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- 10 \pm 4}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- 10 + 4}{2 \cdot 7}, u_{2} = \frac{- 10 - 4}{2 \cdot 7}

u = \frac{- 10 + 4}{2 \cdot 7} : - \frac{3}{7}

\frac{- 10 + 4}{2 \cdot 7}

Add/Subtract the numbers:

- 10 + 4 = - 6

= \frac{- 6}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 6}{14}

Apply the fraction rule:

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

= - \frac{6}{14}

Cancel the common factor:

2

= - \frac{3}{7}

u = \frac{- 10 - 4}{2 \cdot 7} : - 1

\frac{- 10 - 4}{2 \cdot 7}

Subtract the numbers:

- 10 - 4 = - 14

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

Multiply the numbers:

2 \cdot 7 = 14

= \frac{- 14}{14}

Apply the fraction rule:

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

= - \frac{14}{14}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - \frac{3}{7}, u = - 1

Substitute back

u = cos (θ)

cos (θ) = - \frac{3}{7}, cos (θ) = - 1

cos (θ) = - \frac{3}{7}, cos (θ) = - 1

cos (θ) = - \frac{3}{7} : θ = arccos (- \frac{3}{7}) + 2 πn, θ = - arccos (- \frac{3}{7}) + 2 πn

cos (θ) = - \frac{3}{7}

Apply trig inverse properties

cos (θ) = - \frac{3}{7}

General solutions for

cos (θ) = - \frac{3}{7}

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

θ = arccos (- \frac{3}{7}) + 2 πn, θ = - arccos (- \frac{3}{7}) + 2 πn

θ = arccos (- \frac{3}{7}) + 2 πn, θ = - arccos (- \frac{3}{7}) + 2 πn

cos (θ) = - 1 : θ = π + 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 = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = arccos (- \frac{3}{7}) + 2 πn, θ = - arccos (- \frac{3}{7}) + 2 πn, θ = π + 2 πn

Show solutions in decimal form

θ = 2.01370 \dots + 2 πn, θ = - 2.01370 \dots + 2 πn, θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7cos^2(θ)+11cos(θ)=cos(θ)-3 ?
The general solution for 7cos^2(θ)+11cos(θ)=cos(θ)-3 is θ=2.01370…+2pin,θ=-2.01370…+2pin,θ=pi+2pin