What is the general solution for (-2sin(x)-1)/((2+sin(x))^2)=0 ?

The general solution for (-2sin(x)-1)/((2+sin(x))^2)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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(-2sin(x)-1)/((2+sin(x))^2)=0

Solution

\frac{- 2 sin ( x ) - 1}{( 2 + sin ( x ) ) ^{2}} = 0

Solution

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

Degrees

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

Solution steps

\frac{- 2 sin ( x ) - 1}{( 2 + sin ( x ) ) ^{2}} = 0

Solve by substitution

\frac{- 2 sin ( x ) - 1}{( 2 + sin ( x ) ) ^{2}} = 0

Let:

sin (x) = u

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 2 u - 1 = 0

Move

1

to the right side

- 2 u - 1 = 0

Add

1

to both sides

- 2 u - 1 + 1 = 0 + 1

Simplify

- 2 u = 1

- 2 u = 1

Divide both sides by

- 2

- 2 u = 1

Divide both sides by

- 2

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

Simplify

u = - \frac{1}{2}

u = - \frac{1}{2}

Verify Solutions

Find undefined (singularity) points:

u = - 2

Take the denominator(s) of

\frac{- 2 u - 1}{( 2 + u ) ^{2}}

and compare to zero

Solve

(2 + u)^{2} = 0 : u = - 2

(2 + u)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

2 + u = 0 : u = - 2

2 + u = 0

Move

2

to the right side

2 + u = 0

Subtract

2

from both sides

2 + u - 2 = 0 - 2

Simplify

u = - 2

u = - 2

The solution to the quadratic equation is:

u = - 2

The following points are undefined

u = - 2

Combine undefined points with solutions:

u = - \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

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

sin (x) = - \frac{1}{2}

General solutions for

sin (x) = - \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}

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

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

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for (-2sin(x)-1)/((2+sin(x))^2)=0 ?
The general solution for (-2sin(x)-1)/((2+sin(x))^2)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin