What is the general solution for 2cos(2θ)+1=-12cos(θ)-6 ?

The general solution for 2cos(2θ)+1=-12cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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Popular >

2cos(2θ)+1=-12cos(θ)-6

Solution

2 cos (2 θ) + 1 = - 12 cos (θ) - 6

Solution

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (2 θ) + 1 = - 12 cos (θ) - 6

Subtract

- 12 cos (θ) - 6

from both sides

2 cos (2 θ) + 12 cos (θ) + 7 = 0

Rewrite using trig identities

7 + 12 cos (θ) + 2 cos (2 θ)

Use the Double Angle identity:

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

= 7 + 12 cos (θ) + 2 (2 cos^{2} (θ) - 1)

Simplify

7 + 12 cos (θ) + 2 (2 cos^{2} (θ) - 1) : 4 cos^{2} (θ) + 12 cos (θ) + 5

7 + 12 cos (θ) + 2 (2 cos^{2} (θ) - 1)

Expand

2 (2 cos^{2} (θ) - 1) : 4 cos^{2} (θ) - 2

2 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (θ), c = 1

= 2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

Simplify

2 \cdot 2 cos^{2} (θ) - 2 \cdot 1 : 4 cos^{2} (θ) - 2

2 \cdot 2 cos^{2} (θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 2 = 4

= 4 cos^{2} (θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 4 cos^{2} (θ) - 2

= 4 cos^{2} (θ) - 2

= 7 + 12 cos (θ) + 4 cos^{2} (θ) - 2

Simplify

7 + 12 cos (θ) + 4 cos^{2} (θ) - 2 : 4 cos^{2} (θ) + 12 cos (θ) + 5

7 + 12 cos (θ) + 4 cos^{2} (θ) - 2

Group like terms

= 12 cos (θ) + 4 cos^{2} (θ) + 7 - 2

Add/Subtract the numbers:

7 - 2 = 5

= 4 cos^{2} (θ) + 12 cos (θ) + 5

= 4 cos^{2} (θ) + 12 cos (θ) + 5

= 4 cos^{2} (θ) + 12 cos (θ) + 5

5 + 12 cos (θ) + 4 cos^{2} (θ) = 0

Solve by substitution

5 + 12 cos (θ) + 4 cos^{2} (θ) = 0

Let:

cos (θ) = u

5 + 12 u + 4 u^{2} = 0

5 + 12 u + 4 u^{2} = 0 : u = - \frac{1}{2}, u = - \frac{5}{2}

5 + 12 u + 4 u^{2} = 0

Write in the standard form

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

4 u^{2} + 12 u + 5 = 0

Solve with the quadratic formula

4 u^{2} + 12 u + 5 = 0

Quadratic Equation Formula:

For

a = 4, b = 12, c = 5

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

u_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

1 2^{2} - 4 \cdot 4 \cdot 5 = 8

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

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 1 2^{2} - 80

1 2^{2} = 144

= 144 - 80

Subtract the numbers:

144 - 80 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

n a^{n} = a

8^{2} = 8

= 8

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 12 + 8 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

u = \frac{- 12 - 8}{2 \cdot 4} : - \frac{5}{2}

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

Subtract the numbers:

- 12 - 8 = - 20

= \frac{- 20}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 20}{8}

Apply the fraction rule:

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

= - \frac{20}{8}

Cancel the common factor:

4

= - \frac{5}{2}

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = - \frac{5}{2}

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}, cos (θ) = - \frac{5}{2}

cos (θ) = - \frac{1}{2}, cos (θ) = - \frac{5}{2}

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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 = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{5}{2} :

No Solution

cos (θ) = - \frac{5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ)+1=-12cos(θ)-6 ?
The general solution for 2cos(2θ)+1=-12cos(θ)-6 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin