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

The general solution for 3cos^2(θ)+cos(θ)=-2 is No Solution for θ\in\mathbb{R}

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Popular >

3cos^2(θ)+cos(θ)=-2

Solution

3 cos^{2} (θ) + cos (θ) = - 2

Solution

No Solution for θ \in R

Solution steps

3 cos^{2} (θ) + cos (θ) = - 2

Solve by substitution

3 cos^{2} (θ) + cos (θ) = - 2

Let:

cos (θ) = u

3 u^{2} + u = - 2

3 u^{2} + u = - 2 : u = - \frac{1}{6} + i \frac{23}{6}, u = - \frac{1}{6} - i \frac{23}{6}

3 u^{2} + u = - 2

Move

2

to the left side

3 u^{2} + u = - 2

Add

2

to both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 3, b = 1, c = 2

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

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

Simplify

1^{2} - 4 \cdot 3 \cdot 2 : 23 i

1^{2} - 4 \cdot 3 \cdot 2

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 \cdot 2

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 1 - 24

Subtract the numbers:

1 - 24 = - 23

= - 23

Apply radical rule:

- a = - 1 a

- 23 = - 1 23

= - 1 23

Apply imaginary number rule:

- 1 = i

= 23 i

u_{1, 2} = \frac{- 1 \pm 23 i}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 1 + 23 i}{2 \cdot 3}, u_{2} = \frac{- 1 - 23 i}{2 \cdot 3}

u = \frac{- 1 + 23 i}{2 \cdot 3} : - \frac{1}{6} + i \frac{23}{6}

\frac{- 1 + 23 i}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 1 + 23 i}{6}

Rewrite

\frac{- 1 + 23 i}{6}

in standard complex form:

- \frac{1}{6} + \frac{23}{6} i

\frac{- 1 + 23 i}{6}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 23 i}{6} = - \frac{1}{6} + \frac{23 i}{6}

= - \frac{1}{6} + \frac{23 i}{6}

= - \frac{1}{6} + \frac{23}{6} i

u = \frac{- 1 - 23 i}{2 \cdot 3} : - \frac{1}{6} - i \frac{23}{6}

\frac{- 1 - 23 i}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 1 - 23 i}{6}

Rewrite

\frac{- 1 - 23 i}{6}

in standard complex form:

- \frac{1}{6} - \frac{23}{6} i

\frac{- 1 - 23 i}{6}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 23 i}{6} = - \frac{1}{6} - \frac{23 i}{6}

= - \frac{1}{6} - \frac{23 i}{6}

= - \frac{1}{6} - \frac{23}{6} i

The solutions to the quadratic equation are:

u = - \frac{1}{6} + i \frac{23}{6}, u = - \frac{1}{6} - i \frac{23}{6}

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{6} + i \frac{23}{6}, cos (θ) = - \frac{1}{6} - i \frac{23}{6}

cos (θ) = - \frac{1}{6} + i \frac{23}{6}, cos (θ) = - \frac{1}{6} - i \frac{23}{6}

cos (θ) = - \frac{1}{6} + i \frac{23}{6} :

No Solution

cos (θ) = - \frac{1}{6} + i \frac{23}{6}

No Solution

cos (θ) = - \frac{1}{6} - i \frac{23}{6} :

No Solution

cos (θ) = - \frac{1}{6} - i \frac{23}{6}

No Solution

Combine all the solutions

No Solution for θ \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(θ)+cos(θ)=-2 ?
The general solution for 3cos^2(θ)+cos(θ)=-2 is No Solution for θ\in\mathbb{R}