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

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

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Popular >

3/(2cos(θ))=3-3cos(θ)

Solution

\frac{3}{2 cos ( θ )} = 3 - 3 cos (θ)

Solution

No Solution for θ \in R

Solution steps

\frac{3}{2 cos ( θ )} = 3 - 3 cos (θ)

Solve by substitution

\frac{3}{2 cos ( θ )} = 3 - 3 cos (θ)

Let:

cos (θ) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

- 3 uu : - 3 u^{2}

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= - 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 3 u^{2}

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

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

Solve

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

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

Multiply both sides by

2

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

Multiply both sides by

2

\frac{3}{2} \cdot 2 = 3 u \cdot 2 - 3 u^{2} \cdot 2

Simplify

3 = 6 u - 6 u^{2}

3 = 6 u - 6 u^{2}

Switch sides

6 u - 6 u^{2} = 3

Move

3

to the left side

6 u - 6 u^{2} = 3

Subtract

3

from both sides

6 u - 6 u^{2} - 3 = 3 - 3

Simplify

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

6^{2} - 4 (- 6) (- 3) : 6 i

6^{2} - 4 (- 6) (- 3)

Apply rule

- (- a) = a

= 6^{2} - 4 \cdot 6 \cdot 3

Multiply the numbers:

4 \cdot 6 \cdot 3 = 72

= 6^{2} - 72

Apply imaginary number rule:

- a = i a

= i 72 - 6^{2}

- 6^{2} + 72 = 6

- 6^{2} + 72

6^{2} = 36

= - 36 + 72

Add/Subtract the numbers:

- 36 + 72 = 36

= 36

Factor the number:

36 = 6^{2}

= 6^{2}

Apply radical rule:

6^{2} = 6

= 6

= 6 i

u_{1, 2} = \frac{- 6 \pm 6 i}{2 ( - 6 )}

Separate the solutions

u_{1} = \frac{- 6 + 6 i}{2 ( - 6 )}, u_{2} = \frac{- 6 - 6 i}{2 ( - 6 )}

u = \frac{- 6 + 6 i}{2 ( - 6 )} : \frac{1}{2} - i \frac{1}{2}

\frac{- 6 + 6 i}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 6 i}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6 + 6 i}{- 12}

Apply the fraction rule:

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

= - \frac{- 6 + 6 i}{12}

Cancel

\frac{- 6 + 6 i}{12} : \frac{- 1 + i}{2}

\frac{- 6 + 6 i}{12}

Factor

- 6 + 6 i : 6 (- 1 + i)

- 6 + 6 i

Rewrite as

= - 6 \cdot 1 + 6 i

Factor out common term

6

= 6 (- 1 + i)

= \frac{6 ( - 1 + i )}{12}

Cancel the common factor:

6

= \frac{- 1 + i}{2}

= - \frac{- 1 + i}{2}

Rewrite

- \frac{- 1 + i}{2}

in standard complex form:

\frac{1}{2} - \frac{1}{2} i

- \frac{- 1 + i}{2}

Apply the fraction rule:

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

\frac{- 1 + i}{2} = - (- \frac{1}{2}) - (\frac{i}{2})

= - (- \frac{1}{2}) - (\frac{i}{2})

Remove parentheses:

(a) = a, - (- a) = a

= \frac{1}{2} - \frac{i}{2}

= \frac{1}{2} - \frac{1}{2} i

u = \frac{- 6 - 6 i}{2 ( - 6 )} : \frac{1}{2} + i \frac{1}{2}

\frac{- 6 - 6 i}{2 ( - 6 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 6 i}{- 2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{- 6 - 6 i}{- 12}

Apply the fraction rule:

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

= - \frac{- 6 - 6 i}{12}

Cancel

\frac{- 6 - 6 i}{12} : - \frac{1 + i}{2}

\frac{- 6 - 6 i}{12}

Factor

- 6 - 6 i : - 6 (1 + i)

- 6 - 6 i

Rewrite as

= - 6 \cdot 1 - 6 i

Factor out common term

6

= - 6 (1 + i)

= - \frac{6 ( 1 + i )}{12}

Cancel the common factor:

6

= - \frac{1 + i}{2}

= - (- \frac{1 + i}{2})

Apply rule

- (- a) = a

= \frac{1 + i}{2}

Rewrite

\frac{1 + i}{2}

in standard complex form:

\frac{1}{2} + \frac{1}{2} i

\frac{1 + i}{2}

Apply the fraction rule:

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

\frac{1 + i}{2} = \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{1}{2} i

The solutions to the quadratic equation are:

u = \frac{1}{2} - i \frac{1}{2}, u = \frac{1}{2} + i \frac{1}{2}

u = \frac{1}{2} - i \frac{1}{2}, u = \frac{1}{2} + i \frac{1}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2} - i \frac{1}{2}, cos (θ) = \frac{1}{2} + i \frac{1}{2}

cos (θ) = \frac{1}{2} - i \frac{1}{2}, cos (θ) = \frac{1}{2} + i \frac{1}{2}

cos (θ) = \frac{1}{2} - i \frac{1}{2} :

No Solution

cos (θ) = \frac{1}{2} - i \frac{1}{2}

No Solution

cos (θ) = \frac{1}{2} + i \frac{1}{2} :

No Solution

cos (θ) = \frac{1}{2} + i \frac{1}{2}

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