What is the general solution for 3cos^2(θ)+4cos(θ)-4=0 ?

The general solution for 3cos^2(θ)+4cos(θ)-4=0 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin

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Popular >

3cos^2(θ)+4cos(θ)-4=0

Solution

3 cos^{2} (θ) + 4 cos (θ) - 4 = 0

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) + 4 cos (θ) - 4 = 0

Solve by substitution

3 cos^{2} (θ) + 4 cos (θ) - 4 = 0

Let:

cos (θ) = u

3 u^{2} + 4 u - 4 = 0

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

3 u^{2} + 4 u - 4 = 0

Solve with the quadratic formula

3 u^{2} + 4 u - 4 = 0

Quadratic Equation Formula:

For

a = 3, b = 4, c = - 4

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

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

4^{2} - 4 \cdot 3 (- 4) = 8

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 3 \cdot 4 = 48

= 4^{2} + 48

4^{2} = 16

= 16 + 48

Add the numbers:

16 + 48 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- 4 \pm 8}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 4 + 8}{2 \cdot 3}, u_{2} = \frac{- 4 - 8}{2 \cdot 3}

u = \frac{- 4 + 8}{2 \cdot 3} : \frac{2}{3}

\frac{- 4 + 8}{2 \cdot 3}

Add/Subtract the numbers:

- 4 + 8 = 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{- 4 - 8}{2 \cdot 3} : - 2

\frac{- 4 - 8}{2 \cdot 3}

Subtract the numbers:

- 4 - 8 = - 12

= \frac{- 12}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 12}{6}

Apply the fraction rule:

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

= - \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = cos (θ)

cos (θ) = \frac{2}{3}, cos (θ) = - 2

cos (θ) = \frac{2}{3}, cos (θ) = - 2

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) = - 2 :

No Solution

cos (θ) = - 2

- 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)+4cos(θ)-4=0 ?
The general solution for 3cos^2(θ)+4cos(θ)-4=0 is θ=0.84106…+2pin,θ=2pi-0.84106…+2pin