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Popular Trigonometry >

4+4sin(θ)= 3/(1-sin(θ))

Solution

4 + 4 sin (θ) = \frac{3}{1 - sin ( θ )}

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

4 + 4 sin (θ) = \frac{3}{1 - sin ( θ )}

Solve by substitution

4 + 4 sin (θ) = \frac{3}{1 - sin ( θ )}

Let:

sin (θ) = u

4 + 4 u = \frac{3}{1 - u}

4 + 4 u = \frac{3}{1 - u} : u = - \frac{1}{2}, u = \frac{1}{2}

4 + 4 u = \frac{3}{1 - u}

Multiply both sides by

1 - u

4 + 4 u = \frac{3}{1 - u}

Multiply both sides by

1 - u

4 (1 - u) + 4 u (1 - u) = \frac{3}{1 - u} (1 - u)

Simplify

4 (1 - u) + 4 u (1 - u) = 3

4 (1 - u) + 4 u (1 - u) = 3

Solve

4 (1 - u) + 4 u (1 - u) = 3 : u = - \frac{1}{2}, u = \frac{1}{2}

4 (1 - u) + 4 u (1 - u) = 3

Expand

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

4 (1 - u) + 4 u (1 - u)

Expand

4 (1 - u) : 4 - 4 u

4 (1 - u)

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = u

= 4 \cdot 1 - 4 u

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 u

= 4 - 4 u + 4 u (1 - u)

Expand

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

4 u (1 - u)

Apply the distributive law:

a (b - c) = ab - a c

a = 4 u, b = 1, c = u

= 4 u \cdot 1 - 4 uu

= 4 \cdot 1 \cdot u - 4 uu

Simplify

4 \cdot 1 \cdot u - 4 uu : 4 u - 4 u^{2}

4 \cdot 1 \cdot u - 4 uu

4 \cdot 1 \cdot u = 4 u

4 \cdot 1 \cdot u

Multiply the numbers:

4 \cdot 1 = 4

= 4 u

4 uu = 4 u^{2}

4 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 4 u^{2}

= 4 u - 4 u^{2}

= 4 u - 4 u^{2}

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

Add similar elements:

- 4 u + 4 u = 0

= 4 - 4 u^{2}

4 - 4 u^{2} = 3

Move

3

to the left side

4 - 4 u^{2} = 3

Subtract

3

from both sides

4 - 4 u^{2} - 3 = 3 - 3

Simplify

- 4 u^{2} + 1 = 0

- 4 u^{2} + 1 = 0

Solve with the quadratic formula

- 4 u^{2} + 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 0, c = 1

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

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

0^{2} - 4 (- 4) \cdot 1 = 4

0^{2} - 4 (- 4) \cdot 1

Apply rule

0^{a} = 0

0^{2} = 0

= 0 - 4 (- 4) \cdot 1

Apply rule

- (- a) = a

= 0 + 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 0 + 16

Add the numbers:

0 + 16 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- 0 \pm 4}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 0 + 4}{2 ( - 4 )}, u_{2} = \frac{- 0 - 4}{2 ( - 4 )}

u = \frac{- 0 + 4}{2 ( - 4 )} : - \frac{1}{2}

\frac{- 0 + 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 0 + 4 = 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4}{- 8}

Apply the fraction rule:

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

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

u = \frac{- 0 - 4}{2 ( - 4 )} : \frac{1}{2}

\frac{- 0 - 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 0 - 4 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - \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{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

Divide both sides by

- 1

- u = - 1

Divide both sides by

- 1

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

Simplify

u = 1

u = 1

The following points are undefined

u = 1

Combine undefined points with solutions:

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

Substitute back

u = sin (θ)

sin (θ) = - \frac{1}{2}, sin (θ) = \frac{1}{2}

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

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Graph

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