What is the general solution for 2cos(2θ)+5cos(θ)=4 ?

The general solution for 2cos(2θ)+5cos(θ)=4 is θ=0.72273…+2pin,θ=2pi-0.72273…+2pin

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2cos(2θ)+5cos(θ)=4

Solution

2 cos (2 θ) + 5 cos (θ) = 4

Solution

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

Degrees

θ = 41.40962 \dots^{\circ} + 36 0^{\circ} n, θ = 318.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (2 θ) + 5 cos (θ) = 4

Subtract

4

from both sides

2 cos (2 θ) + 5 cos (θ) - 4 = 0

Rewrite using trig identities

- 4 + 2 cos (2 θ) + 5 cos (θ)

Use the Double Angle identity:

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

= - 4 + 2 (2 cos^{2} (θ) - 1) + 5 cos (θ)

Simplify

- 4 + 2 (2 cos^{2} (θ) - 1) + 5 cos (θ) : 4 cos^{2} (θ) + 5 cos (θ) - 6

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

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

= - 4 + 4 cos^{2} (θ) - 2 + 5 cos (θ)

Simplify

- 4 + 4 cos^{2} (θ) - 2 + 5 cos (θ) : 4 cos^{2} (θ) + 5 cos (θ) - 6

- 4 + 4 cos^{2} (θ) - 2 + 5 cos (θ)

Group like terms

= 4 cos^{2} (θ) + 5 cos (θ) - 4 - 2

Subtract the numbers:

- 4 - 2 = - 6

= 4 cos^{2} (θ) + 5 cos (θ) - 6

= 4 cos^{2} (θ) + 5 cos (θ) - 6

= 4 cos^{2} (θ) + 5 cos (θ) - 6

- 6 + 4 cos^{2} (θ) + 5 cos (θ) = 0

Solve by substitution

- 6 + 4 cos^{2} (θ) + 5 cos (θ) = 0

Let:

cos (θ) = u

- 6 + 4 u^{2} + 5 u = 0

- 6 + 4 u^{2} + 5 u = 0 : u = \frac{3}{4}, u = - 2

- 6 + 4 u^{2} + 5 u = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = 5, c = - 6

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

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

5^{2} - 4 \cdot 4 (- 6) = 11

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 4 \cdot 6 = 96

= 5^{2} + 96

5^{2} = 25

= 25 + 96

Add the numbers:

25 + 96 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 5 \pm 11}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- 5 + 11}{2 \cdot 4}, u_{2} = \frac{- 5 - 11}{2 \cdot 4}

u = \frac{- 5 + 11}{2 \cdot 4} : \frac{3}{4}

\frac{- 5 + 11}{2 \cdot 4}

Add/Subtract the numbers:

- 5 + 11 = 6

= \frac{6}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{6}{8}

Cancel the common factor:

2

= \frac{3}{4}

u = \frac{- 5 - 11}{2 \cdot 4} : - 2

\frac{- 5 - 11}{2 \cdot 4}

Subtract the numbers:

- 5 - 11 = - 16

= \frac{- 16}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 16}{8}

Apply the fraction rule:

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

= - \frac{16}{8}

Divide the numbers:

\frac{16}{8} = 2

= - 2

The solutions to the quadratic equation are:

u = \frac{3}{4}, u = - 2

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{4}, cos (θ) = - 2

cos (θ) = \frac{3}{4}, cos (θ) = - 2

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

cos (θ) = \frac{3}{4}

Apply trig inverse properties

cos (θ) = \frac{3}{4}

General solutions for

cos (θ) = \frac{3}{4}

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

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

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

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Show solutions in decimal form

θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ)+5cos(θ)=4 ?
The general solution for 2cos(2θ)+5cos(θ)=4 is θ=0.72273…+2pin,θ=2pi-0.72273…+2pin