What is the general solution for -5sin^2(θ)+17cos(θ)+9=4cos(θ)-2 ?

The general solution for -5sin^2(θ)+17cos(θ)+9=4cos(θ)-2 is θ=2.21429…+2pin,θ=-2.21429…+2pin

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-5sin^2(θ)+17cos(θ)+9=4cos(θ)-2

Solution

- 5 sin^{2} (θ) + 17 cos (θ) + 9 = 4 cos (θ) - 2

Solution

θ = 2.21429 \dots + 2 πn, θ = - 2.21429 \dots + 2 πn

Degrees

θ = 126.86989 \dots^{\circ} + 36 0^{\circ} n, θ = - 126.86989 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 5 sin^{2} (θ) + 17 cos (θ) + 9 = 4 cos (θ) - 2

Subtract

4 cos (θ) - 2

from both sides

13 cos (θ) - 5 sin^{2} (θ) + 11 = 0

Rewrite using trig identities

11 + 13 cos (θ) - 5 sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= 11 + 13 cos (θ) - 5 (1 - cos^{2} (θ))

Simplify

11 + 13 cos (θ) - 5 (1 - cos^{2} (θ)) : 5 cos^{2} (θ) + 13 cos (θ) + 6

11 + 13 cos (θ) - 5 (1 - cos^{2} (θ))

Expand

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

- 5 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 5, b = 1, c = cos^{2} (θ)

= - 5 \cdot 1 - (- 5) cos^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 5 \cdot 1 + 5 cos^{2} (θ)

Multiply the numbers:

5 \cdot 1 = 5

= - 5 + 5 cos^{2} (θ)

= 11 + 13 cos (θ) - 5 + 5 cos^{2} (θ)

Simplify

11 + 13 cos (θ) - 5 + 5 cos^{2} (θ) : 5 cos^{2} (θ) + 13 cos (θ) + 6

11 + 13 cos (θ) - 5 + 5 cos^{2} (θ)

Group like terms

= 13 cos (θ) + 5 cos^{2} (θ) + 11 - 5

Add/Subtract the numbers:

11 - 5 = 6

= 5 cos^{2} (θ) + 13 cos (θ) + 6

= 5 cos^{2} (θ) + 13 cos (θ) + 6

= 5 cos^{2} (θ) + 13 cos (θ) + 6

6 + 13 cos (θ) + 5 cos^{2} (θ) = 0

Solve by substitution

6 + 13 cos (θ) + 5 cos^{2} (θ) = 0

Let:

cos (θ) = u

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

6 + 13 u + 5 u^{2} = 0 : u = - \frac{3}{5}, u = - 2

6 + 13 u + 5 u^{2} = 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 \cdot 6}{2 \cdot 5}

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

1 3^{2} - 4 \cdot 5 \cdot 6 = 7

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

Subtract the numbers:

169 - 120 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 13 \pm 7}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- 13 + 7}{2 \cdot 5}, u_{2} = \frac{- 13 - 7}{2 \cdot 5}

u = \frac{- 13 + 7}{2 \cdot 5} : - \frac{3}{5}

\frac{- 13 + 7}{2 \cdot 5}

Add/Subtract the numbers:

- 13 + 7 = - 6

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

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 6}{10}

Apply the fraction rule:

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

= - \frac{6}{10}

Cancel the common factor:

2

= - \frac{3}{5}

u = \frac{- 13 - 7}{2 \cdot 5} : - 2

\frac{- 13 - 7}{2 \cdot 5}

Subtract the numbers:

- 13 - 7 = - 20

= \frac{- 20}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 20}{10}

Apply the fraction rule:

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

= - \frac{20}{10}

Divide the numbers:

\frac{20}{10} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = - \frac{3}{5}, cos (θ) = - 2

cos (θ) = - \frac{3}{5}, cos (θ) = - 2

cos (θ) = - \frac{3}{5} : θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

cos (θ) = - \frac{3}{5}

Apply trig inverse properties

cos (θ) = - \frac{3}{5}

General solutions for

cos (θ) = - \frac{3}{5}

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

θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

θ = arccos (- \frac{3}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 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}{5}) + 2 πn, θ = - arccos (- \frac{3}{5}) + 2 πn

Show solutions in decimal form

θ = 2.21429 \dots + 2 πn, θ = - 2.21429 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -5sin^2(θ)+17cos(θ)+9=4cos(θ)-2 ?
The general solution for -5sin^2(θ)+17cos(θ)+9=4cos(θ)-2 is θ=2.21429…+2pin,θ=-2.21429…+2pin