What is the general solution for (1-sin(θ))/(3+3sin(θ))=1 ?

The general solution for (1-sin(θ))/(3+3sin(θ))=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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Popular >

(1-sin(θ))/(3+3sin(θ))=1

Solution

\frac{1 - sin ( θ )}{3 + 3 sin ( θ )} = 1

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{1 - sin ( θ )}{3 + 3 sin ( θ )} = 1

Solve by substitution

\frac{1 - sin ( θ )}{3 + 3 sin ( θ )} = 1

Let:

sin (θ) = u

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

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

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

Multiply both sides by

3 + 3 u

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

Multiply both sides by

3 + 3 u

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

Simplify

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

3 + 3 u

= 1 - u

Simplify

1 \cdot (3 + 3 u) : 3 + 3 u

1 \cdot (3 + 3 u)

Multiply:

1 \cdot (3 + 3 u) = (3 + 3 u)

= (3 + 3 u)

Remove parentheses:

(a) = a

= 3 + 3 u

1 - u = 3 + 3 u

1 - u = 3 + 3 u

1 - u = 3 + 3 u

Move

1

to the right side

1 - u = 3 + 3 u

Subtract

1

from both sides

1 - u - 1 = 3 + 3 u - 1

Simplify

- u = 3 u + 2

- u = 3 u + 2

Move

3 u

to the left side

- u = 3 u + 2

Subtract

3 u

from both sides

- u - 3 u = 3 u + 2 - 3 u

Simplify

- 4 u = 2

- 4 u = 2

Divide both sides by

- 4

- 4 u = 2

Divide both sides by

- 4

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

Simplify

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

Simplify

\frac{- 4 u}{- 4} : u

\frac{- 4 u}{- 4}

Apply the fraction rule:

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

= \frac{4 u}{4}

Divide the numbers:

\frac{4}{4} = 1

= u

Simplify

\frac{2}{- 4} : - \frac{1}{2}

\frac{2}{- 4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = - \frac{1}{2}

u = - \frac{1}{2}

u = - \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

u = - 1

Take the denominator(s) of

\frac{1 - u}{3 + 3 u}

and compare to zero

Solve

3 + 3 u = 0 : u = - 1

3 + 3 u = 0

Move

3

to the right side

3 + 3 u = 0

Subtract

3

from both sides

3 + 3 u - 3 = 0 - 3

Simplify

3 u = - 3

3 u = - 3

Divide both sides by

3

3 u = - 3

Divide both sides by

3

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

Simplify

u = - 1

u = - 1

The following points are undefined

u = - 1

Combine undefined points with solutions:

u = - \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Popular Examples

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cos (θ) = \frac{12}{15}

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arctan (x) = 45

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10sin(t)cos(t)=-6sin(t)

10 sin (t) cos (t) = - 6 sin (t)

Frequently Asked Questions (FAQ)

What is the general solution for (1-sin(θ))/(3+3sin(θ))=1 ?
The general solution for (1-sin(θ))/(3+3sin(θ))=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin