What is the general solution for 5-6cos(θ)=3sqrt(3)+5-3cos(θ) ?

The general solution for 5-6cos(θ)=3sqrt(3)+5-3cos(θ) is No Solution for θ\in\mathbb{R}

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Popular >

5-6cos(θ)=3sqrt(3)+5-3cos(θ)

Solution

5 - 6 cos (θ) = 33 + 5 - 3 cos (θ)

Solution

No Solution for θ \in R

Solution steps

5 - 6 cos (θ) = 33 + 5 - 3 cos (θ)

Solve by substitution

5 - 6 cos (θ) = 33 + 5 - 3 cos (θ)

Let:

cos (θ) = u

5 - 6 u = 33 + 5 - 3 u

5 - 6 u = 33 + 5 - 3 u : u = - 3

5 - 6 u = 33 + 5 - 3 u

Subtract

5

from both sides

5 - 6 u - 5 = 33 + 5 - 3 u - 5

Simplify

- 6 u = 33 - 3 u

Move

3 u

to the left side

- 6 u = 33 - 3 u

Add

3 u

to both sides

- 6 u + 3 u = 33 - 3 u + 3 u

Simplify

- 3 u = 33

- 3 u = 33

Divide both sides by

- 3

- 3 u = 33

Divide both sides by

- 3

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

Simplify

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

Simplify

\frac{- 3 u}{- 3} : u

\frac{- 3 u}{- 3}

Apply the fraction rule:

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

= \frac{3 u}{3}

Divide the numbers:

\frac{3}{3} = 1

= u

Simplify

\frac{3 3}{- 3} : - 3

\frac{3 3}{- 3}

Apply the fraction rule:

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

= - \frac{3 3}{3}

Divide the numbers:

\frac{3}{3} = 1

= - 3

u = - 3

u = - 3

u = - 3

Substitute back

u = cos (θ)

cos (θ) = - 3

cos (θ) = - 3

cos (θ) = - 3 :

No Solution

cos (θ) = - 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 5-6cos(θ)=3sqrt(3)+5-3cos(θ) ?
The general solution for 5-6cos(θ)=3sqrt(3)+5-3cos(θ) is No Solution for θ\in\mathbb{R}