What is the general solution for 3cos(2θ)+5cos(θ)+2=0 ?

The general solution for 3cos(2θ)+5cos(θ)+2=0 is θ=1.40334…+2pin,θ=2pi-1.40334…+2pin,θ=pi+2pin

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3cos(2θ)+5cos(θ)+2=0

Solution

3 cos (2 θ) + 5 cos (θ) + 2 = 0

Solution

θ = 1.40334 \dots + 2 πn, θ = 2 π - 1.40334 \dots + 2 πn, θ = π + 2 πn

Degrees

θ = 80.40593 \dots^{\circ} + 36 0^{\circ} n, θ = 279.59406 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos (2 θ) + 5 cos (θ) + 2 = 0

Rewrite using trig identities

2 + 3 cos (2 θ) + 5 cos (θ)

Use the Double Angle identity:

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

= 2 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ)

Simplify

2 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ) : 6 cos^{2} (θ) + 5 cos (θ) - 1

2 + 3 (2 cos^{2} (θ) - 1) + 5 cos (θ)

Expand

3 (2 cos^{2} (θ) - 1) : 6 cos^{2} (θ) - 3

3 (2 cos^{2} (θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 2 cos^{2} (θ), c = 1

= 3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

Simplify

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1 : 6 cos^{2} (θ) - 3

3 \cdot 2 cos^{2} (θ) - 3 \cdot 1

Multiply the numbers:

3 \cdot 2 = 6

= 6 cos^{2} (θ) - 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 6 cos^{2} (θ) - 3

= 6 cos^{2} (θ) - 3

= 2 + 6 cos^{2} (θ) - 3 + 5 cos (θ)

Simplify

2 + 6 cos^{2} (θ) - 3 + 5 cos (θ) : 6 cos^{2} (θ) + 5 cos (θ) - 1

2 + 6 cos^{2} (θ) - 3 + 5 cos (θ)

Group like terms

= 6 cos^{2} (θ) + 5 cos (θ) + 2 - 3

Add/Subtract the numbers:

2 - 3 = - 1

= 6 cos^{2} (θ) + 5 cos (θ) - 1

= 6 cos^{2} (θ) + 5 cos (θ) - 1

= 6 cos^{2} (θ) + 5 cos (θ) - 1

- 1 + 5 cos (θ) + 6 cos^{2} (θ) = 0

Solve by substitution

- 1 + 5 cos (θ) + 6 cos^{2} (θ) = 0

Let:

cos (θ) = u

- 1 + 5 u + 6 u^{2} = 0

- 1 + 5 u + 6 u^{2} = 0 : u = \frac{1}{6}, u = - 1

- 1 + 5 u + 6 u^{2} = 0

Write in the standard form

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

6 u^{2} + 5 u - 1 = 0

Solve with the quadratic formula

6 u^{2} + 5 u - 1 = 0

Quadratic Equation Formula:

For

a = 6, b = 5, c = - 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 6 ( - 1 )}{2 \cdot 6}

5^{2} - 4 \cdot 6 (- 1) = 7

5^{2} - 4 \cdot 6 (- 1)

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- 5 + 7}{2 \cdot 6}, u_{2} = \frac{- 5 - 7}{2 \cdot 6}

u = \frac{- 5 + 7}{2 \cdot 6} : \frac{1}{6}

\frac{- 5 + 7}{2 \cdot 6}

Add/Subtract the numbers:

- 5 + 7 = 2

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{2}{12}

Cancel the common factor:

2

= \frac{1}{6}

u = \frac{- 5 - 7}{2 \cdot 6} : - 1

\frac{- 5 - 7}{2 \cdot 6}

Subtract the numbers:

- 5 - 7 = - 12

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 12}{12}

Apply the fraction rule:

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

= - \frac{12}{12}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{6}, cos (θ) = - 1

cos (θ) = \frac{1}{6}, cos (θ) = - 1

cos (θ) = \frac{1}{6} : θ = arccos (\frac{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 2 πn

cos (θ) = \frac{1}{6}

Apply trig inverse properties

cos (θ) = \frac{1}{6}

General solutions for

cos (θ) = \frac{1}{6}

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

θ = arccos (\frac{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 2 πn

θ = arccos (\frac{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 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{1}{6}) + 2 πn, θ = 2 π - arccos (\frac{1}{6}) + 2 πn, θ = π + 2 πn

Show solutions in decimal form

θ = 1.40334 \dots + 2 πn, θ = 2 π - 1.40334 \dots + 2 πn, θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos(2θ)+5cos(θ)+2=0 ?
The general solution for 3cos(2θ)+5cos(θ)+2=0 is θ=1.40334…+2pin,θ=2pi-1.40334…+2pin,θ=pi+2pin