What is the general solution for 5cos^2(θ)=-13cos(θ)+6 ?

The general solution for 5cos^2(θ)=-13cos(θ)+6 is θ=1.15927…+2pin,θ=2pi-1.15927…+2pin

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Popular >

5cos^2(θ)=-13cos(θ)+6

Solution

5 cos^{2} (θ) = - 13 cos (θ) + 6

Solution

θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn

Degrees

θ = 66.42182 \dots^{\circ} + 36 0^{\circ} n, θ = 293.57817 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 cos^{2} (θ) = - 13 cos (θ) + 6

Solve by substitution

5 cos^{2} (θ) = - 13 cos (θ) + 6

Let:

cos (θ) = u

5 u^{2} = - 13 u + 6

5 u^{2} = - 13 u + 6 : u = \frac{2}{5}, u = - 3

5 u^{2} = - 13 u + 6

Move

6

to the left side

5 u^{2} = - 13 u + 6

Subtract

6

from both sides

5 u^{2} - 6 = - 13 u + 6 - 6

Simplify

5 u^{2} - 6 = - 13 u

5 u^{2} - 6 = - 13 u

Move

13 u

to the left side

5 u^{2} - 6 = - 13 u

Add

13 u

to both sides

5 u^{2} - 6 + 13 u = - 13 u + 13 u

Simplify

5 u^{2} - 6 + 13 u = 0

5 u^{2} - 6 + 13 u = 0

Write in the standard form

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

5 u^{2} + 13 u - 6 = 0

Solve with the quadratic formula

5 u^{2} + 13 u - 6 = 0

Quadratic Equation Formula:

For

a = 5, b = 13, c = - 6

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 5 ( - 6 )}{2 \cdot 5}

u_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 5 ( - 6 )}{2 \cdot 5}

1 3^{2} - 4 \cdot 5 (- 6) = 17

1 3^{2} - 4 \cdot 5 (- 6)

Apply rule

- (- a) = a

= 1 3^{2} + 4 \cdot 5 \cdot 6

Multiply the numbers:

4 \cdot 5 \cdot 6 = 120

= 1 3^{2} + 120

1 3^{2} = 169

= 169 + 120

Add the numbers:

169 + 120 = 289

= 289

Factor the number:

289 = 1 7^{2}

= 1 7^{2}

Apply radical rule:

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- 13 \pm 17}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 13 + 17}{2 \cdot 5}, u_{2} = \frac{- 13 - 17}{2 \cdot 5}

u = \frac{- 13 + 17}{2 \cdot 5} : \frac{2}{5}

\frac{- 13 + 17}{2 \cdot 5}

Add/Subtract the numbers:

- 13 + 17 = 4

= \frac{4}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{4}{10}

Cancel the common factor:

2

= \frac{2}{5}

u = \frac{- 13 - 17}{2 \cdot 5} : - 3

\frac{- 13 - 17}{2 \cdot 5}

Subtract the numbers:

- 13 - 17 = - 30

= \frac{- 30}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 30}{10}

Apply the fraction rule:

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

= - \frac{30}{10}

Divide the numbers:

\frac{30}{10} = 3

= - 3

The solutions to the quadratic equation are:

u = \frac{2}{5}, u = - 3

Substitute back

u = cos (θ)

cos (θ) = \frac{2}{5}, cos (θ) = - 3

cos (θ) = \frac{2}{5}, cos (θ) = - 3

cos (θ) = \frac{2}{5} : θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

cos (θ) = \frac{2}{5}

Apply trig inverse properties

cos (θ) = \frac{2}{5}

General solutions for

cos (θ) = \frac{2}{5}

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

θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

cos (θ) = - 3 :

No Solution

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{2}{5}) + 2 πn, θ = 2 π - arccos (\frac{2}{5}) + 2 πn

Show solutions in decimal form

θ = 1.15927 \dots + 2 πn, θ = 2 π - 1.15927 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5cos^2(θ)=-13cos(θ)+6 ?
The general solution for 5cos^2(θ)=-13cos(θ)+6 is θ=1.15927…+2pin,θ=2pi-1.15927…+2pin