What is the general solution for 3/(1+cos(θ))=1+2cos(θ) ?

The general solution for 3/(1+cos(θ))=1+2cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin

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3/(1+cos(θ))=1+2cos(θ)

Solution

\frac{3}{1 + cos ( θ )} = 1 + 2 cos (θ)

Solution

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{3}{1 + cos ( θ )} = 1 + 2 cos (θ)

Solve by substitution

\frac{3}{1 + cos ( θ )} = 1 + 2 cos (θ)

Let:

cos (θ) = u

\frac{3}{1 + u} = 1 + 2 u

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

\frac{3}{1 + u} = 1 + 2 u

Multiply both sides by

1 + u

\frac{3}{1 + u} = 1 + 2 u

Multiply both sides by

1 + u

\frac{3}{1 + u} (1 + u) = 1 \cdot (1 + u) + 2 u (1 + u)

Simplify

\frac{3}{1 + u} (1 + u) = 1 \cdot (1 + u) + 2 u (1 + u)

Simplify

\frac{3}{1 + u} (1 + u) : 3

\frac{3}{1 + u} (1 + u)

Multiply fractions:

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

= \frac{3 ( 1 + u )}{1 + u}

Cancel the common factor:

1 + u

= 3

Simplify

1 \cdot (1 + u) : 1 + u

1 \cdot (1 + u)

Multiply:

1 \cdot (1 + u) = (1 + u)

= (1 + u)

Remove parentheses:

(a) = a

= 1 + u

3 = 1 + u + 2 u (1 + u)

3 = 1 + u + 2 u (1 + u)

3 = 1 + u + 2 u (1 + u)

Solve

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

3 = 1 + u + 2 u (1 + u)

Expand

1 + u + 2 u (1 + u) : 1 + 3 u + 2 u^{2}

1 + u + 2 u (1 + u)

Expand

2 u (1 + u) : 2 u + 2 u^{2}

2 u (1 + u)

Apply the distributive law:

a (b + c) = ab + a c

a = 2 u, b = 1, c = u

= 2 u \cdot 1 + 2 uu

= 2 \cdot 1 \cdot u + 2 uu

Simplify

2 \cdot 1 \cdot u + 2 uu : 2 u + 2 u^{2}

2 \cdot 1 \cdot u + 2 uu

2 \cdot 1 \cdot u = 2 u

2 \cdot 1 \cdot u

Multiply the numbers:

2 \cdot 1 = 2

= 2 u

2 uu = 2 u^{2}

2 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 2 u^{2}

= 2 u + 2 u^{2}

= 2 u + 2 u^{2}

= 1 + u + 2 u + 2 u^{2}

Add similar elements:

u + 2 u = 3 u

= 1 + 3 u + 2 u^{2}

3 = 1 + 3 u + 2 u^{2}

Switch sides

1 + 3 u + 2 u^{2} = 3

Move

3

to the left side

1 + 3 u + 2 u^{2} = 3

Subtract

3

from both sides

1 + 3 u + 2 u^{2} - 3 = 3 - 3

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = 3, c = - 2

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

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

3^{2} - 4 \cdot 2 (- 2) = 5

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 5 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

Subtract the numbers:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

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

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

Verify Solutions

Find undefined (singularity) points:

u = - 1

Take the denominator(s) of

\frac{3}{1 + u}

and compare to zero

Solve

1 + u = 0 : u = - 1

1 + u = 0

Move

1

to the right side

1 + u = 0

Subtract

1

from both sides

1 + u - 1 = 0 - 1

Simplify

u = - 1

u = - 1

The following points are undefined

u = - 1

Combine undefined points with solutions:

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

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2}, cos (θ) = - 2

cos (θ) = \frac{1}{2}, cos (θ) = - 2

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3/(1+cos(θ))=1+2cos(θ) ?
The general solution for 3/(1+cos(θ))=1+2cos(θ) is θ= pi/3+2pin,θ=(5pi)/3+2pin