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

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

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2+2cos(θ)= 2/(1+cos(θ))

Solution

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

Solution

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (θ) = u

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

2 + 2 u = \frac{2}{1 + u} : u = 0, u = - 2

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

Multiply both sides by

1 + u

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

Multiply both sides by

1 + u

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

Simplify

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

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

Solve

2 (1 + u) + 2 u (1 + u) = 2 : u = 0, u = - 2

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

Expand

2 (1 + u) + 2 u (1 + u) : 2 + 4 u + 2 u^{2}

2 (1 + u) + 2 u (1 + u)

Expand

2 (1 + u) : 2 + 2 u

2 (1 + u)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = 1, c = u

= 2 \cdot 1 + 2 u

Multiply the numbers:

2 \cdot 1 = 2

= 2 + 2 u

= 2 + 2 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}

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

Add similar elements:

2 u + 2 u = 4 u

= 2 + 4 u + 2 u^{2}

2 + 4 u + 2 u^{2} = 2

Move

2

to the left side

2 + 4 u + 2 u^{2} = 2

Subtract

2

from both sides

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

Simplify

2 u^{2} + 4 u = 0

2 u^{2} + 4 u = 0

Solve with the quadratic formula

2 u^{2} + 4 u = 0

Quadratic Equation Formula:

For

a = 2, b = 4, c = 0

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

4^{2} - 4 \cdot 2 \cdot 0 = 4

4^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 4^{2} - 0

4^{2} - 0 = 4^{2}

= 4^{2}

Apply radical rule:

assuming

a \geq 0

= 4

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

Separate the solutions

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

u = \frac{- 4 + 4}{2 \cdot 2} : 0

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

Add/Subtract the numbers:

- 4 + 4 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 4 - 4}{2 \cdot 2} : - 2

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

Subtract the numbers:

- 4 - 4 = - 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 = 0, u = - 2

u = 0, u = - 2

Verify Solutions

Find undefined (singularity) points:

u = - 1

Take the denominator(s) of

\frac{2}{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 = 0, u = - 2

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = - 2

cos (θ) = 0, cos (θ) = - 2

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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

Popular Examples

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14 sin (2 x) + 7 = 0

tan(3x)+sec(3x)=2

tan (3 x) + sec (3 x) = 2

cos^2(x)+4cos(x)-2=0

cos^{2} (x) + 4 cos (x) - 2 = 0

sin(θ)=(-2)/3

sin (θ) = \frac{- 2}{3}

sec^2(θ)+csc^2(θ)=4

sec^{2} (θ) + csc^{2} (θ) = 4

Frequently Asked Questions (FAQ)

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