What is the general solution for 9cos^2(θ)=-10cos(θ)-1 ?

The general solution for 9cos^2(θ)=-10cos(θ)-1 is θ=1.68213…+2pin,θ=-1.68213…+2pin,θ=pi+2pin

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9cos^2(θ)=-10cos(θ)-1

Solution

9 cos^{2} (θ) = - 10 cos (θ) - 1

Solution

θ = 1.68213 \dots + 2 πn, θ = - 1.68213 \dots + 2 πn, θ = π + 2 πn

Degrees

θ = 96.37937 \dots^{\circ} + 36 0^{\circ} n, θ = - 96.37937 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

9 cos^{2} (θ) = - 10 cos (θ) - 1

Solve by substitution

9 cos^{2} (θ) = - 10 cos (θ) - 1

Let:

cos (θ) = u

9 u^{2} = - 10 u - 1

9 u^{2} = - 10 u - 1 : u = - \frac{1}{9}, u = - 1

9 u^{2} = - 10 u - 1

Move

1

to the left side

9 u^{2} = - 10 u - 1

Add

1

to both sides

9 u^{2} + 1 = - 10 u - 1 + 1

Simplify

9 u^{2} + 1 = - 10 u

9 u^{2} + 1 = - 10 u

Move

10 u

to the left side

9 u^{2} + 1 = - 10 u

Add

10 u

to both sides

9 u^{2} + 1 + 10 u = - 10 u + 10 u

Simplify

9 u^{2} + 1 + 10 u = 0

9 u^{2} + 1 + 10 u = 0

Write in the standard form

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

9 u^{2} + 10 u + 1 = 0

Solve with the quadratic formula

9 u^{2} + 10 u + 1 = 0

Quadratic Equation Formula:

For

a = 9, b = 10, c = 1

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

1 0^{2} - 4 \cdot 9 \cdot 1 = 8

1 0^{2} - 4 \cdot 9 \cdot 1

Multiply the numbers:

4 \cdot 9 \cdot 1 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

Subtract the numbers:

100 - 36 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 10 \pm 8}{2 \cdot 9}

Separate the solutions

u_{1} = \frac{- 10 + 8}{2 \cdot 9}, u_{2} = \frac{- 10 - 8}{2 \cdot 9}

u = \frac{- 10 + 8}{2 \cdot 9} : - \frac{1}{9}

\frac{- 10 + 8}{2 \cdot 9}

Add/Subtract the numbers:

- 10 + 8 = - 2

= \frac{- 2}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 2}{18}

Apply the fraction rule:

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

= - \frac{2}{18}

Cancel the common factor:

2

= - \frac{1}{9}

u = \frac{- 10 - 8}{2 \cdot 9} : - 1

\frac{- 10 - 8}{2 \cdot 9}

Subtract the numbers:

- 10 - 8 = - 18

= \frac{- 18}{2 \cdot 9}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{- 18}{18}

Apply the fraction rule:

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

= - \frac{18}{18}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = - \frac{1}{9}, u = - 1

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{9}, cos (θ) = - 1

cos (θ) = - \frac{1}{9}, cos (θ) = - 1

cos (θ) = - \frac{1}{9} : θ = arccos (- \frac{1}{9}) + 2 πn, θ = - arccos (- \frac{1}{9}) + 2 πn

cos (θ) = - \frac{1}{9}

Apply trig inverse properties

cos (θ) = - \frac{1}{9}

General solutions for

cos (θ) = - \frac{1}{9}

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

θ = arccos (- \frac{1}{9}) + 2 πn, θ = - arccos (- \frac{1}{9}) + 2 πn

θ = arccos (- \frac{1}{9}) + 2 πn, θ = - arccos (- \frac{1}{9}) + 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}{9}) + 2 πn, θ = - arccos (- \frac{1}{9}) + 2 πn, θ = π + 2 πn

Show solutions in decimal form

θ = 1.68213 \dots + 2 πn, θ = - 1.68213 \dots + 2 πn, θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 9cos^2(θ)=-10cos(θ)-1 ?
The general solution for 9cos^2(θ)=-10cos(θ)-1 is θ=1.68213…+2pin,θ=-1.68213…+2pin,θ=pi+2pin