What is the general solution for 3cos^2(θ)+7cos(θ)-9=-3 ?

The general solution for 3cos^2(θ)+7cos(θ)-9=-3 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin

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3cos^2(θ)+7cos(θ)-9=-3

Solution

3 cos^{2} (θ) + 7 cos (θ) - 9 = - 3

Solution

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

Degrees

θ = 48.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 311.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (θ) + 7 cos (θ) - 9 = - 3

Solve by substitution

3 cos^{2} (θ) + 7 cos (θ) - 9 = - 3

Let:

cos (θ) = u

3 u^{2} + 7 u - 9 = - 3

3 u^{2} + 7 u - 9 = - 3 : u = \frac{2}{3}, u = - 3

3 u^{2} + 7 u - 9 = - 3

Move

3

to the left side

3 u^{2} + 7 u - 9 = - 3

Add

3

to both sides

3 u^{2} + 7 u - 9 + 3 = - 3 + 3

Simplify

3 u^{2} + 7 u - 6 = 0

3 u^{2} + 7 u - 6 = 0

Solve with the quadratic formula

3 u^{2} + 7 u - 6 = 0

Quadratic Equation Formula:

For

a = 3, b = 7, c = - 6

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

7^{2} - 4 \cdot 3 (- 6) = 11

7^{2} - 4 \cdot 3 (- 6)

Apply rule

- (- a) = a

= 7^{2} + 4 \cdot 3 \cdot 6

Multiply the numbers:

4 \cdot 3 \cdot 6 = 72

= 7^{2} + 72

7^{2} = 49

= 49 + 72

Add the numbers:

49 + 72 = 121

= 121

Factor the number:

121 = 1 1^{2}

= 1 1^{2}

Apply radical rule:

1 1^{2} = 11

= 11

u_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 7 + 11}{2 \cdot 3}, u_{2} = \frac{- 7 - 11}{2 \cdot 3}

u = \frac{- 7 + 11}{2 \cdot 3} : \frac{2}{3}

\frac{- 7 + 11}{2 \cdot 3}

Add/Subtract the numbers:

- 7 + 11 = 4

= \frac{4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

u = \frac{- 7 - 11}{2 \cdot 3} : - 3

\frac{- 7 - 11}{2 \cdot 3}

Subtract the numbers:

- 7 - 11 = - 18

= \frac{- 18}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 18}{6}

Apply the fraction rule:

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

= - \frac{18}{6}

Divide the numbers:

\frac{18}{6} = 3

= - 3

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = \frac{2}{3}, cos (θ) = - 3

cos (θ) = \frac{2}{3}, cos (θ) = - 3

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

cos (θ) = \frac{2}{3}

Apply trig inverse properties

cos (θ) = \frac{2}{3}

General solutions for

cos (θ) = \frac{2}{3}

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

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

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

Show solutions in decimal form

θ = 0.84106 \dots + 2 πn, θ = 2 π - 0.84106 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(θ)+7cos(θ)-9=-3 ?
The general solution for 3cos^2(θ)+7cos(θ)-9=-3 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin