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

The general solution for 1/((2-sin(x)))=sin(x) is x= pi/2+2pin

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1/((2-sin(x)))=sin(x)

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Multiply both sides by

2 - u

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

Multiply both sides by

2 - u

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

Simplify

1 = u (2 - u)

1 = u (2 - u)

Solve

1 = u (2 - u) : u = 1

1 = u (2 - u)

Expand

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

u (2 - u)

Apply the distributive law:

a (b - c) = ab - a c

a = u, b = 2, c = u

= u \cdot 2 - uu

= 2 u - uu

uu = u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

= 2 u - u^{2}

1 = 2 u - u^{2}

Switch sides

2 u - u^{2} = 1

Move

1

to the left side

2 u - u^{2} = 1

Subtract

1

from both sides

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

Simplify

2 u - u^{2} - 1 = 0

2 u - u^{2} - 1 = 0

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} + 2 u - 1 = 0

Solve with the quadratic formula

- u^{2} + 2 u - 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = 2, c = - 1

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

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

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

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

Apply rule

- (- a) = a

= 2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

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

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

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

\frac{- 2}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

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

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

u = 1

Verify Solutions

Find undefined (singularity) points:

u = 2

Take the denominator(s) of

\frac{1}{2 - u}

and compare to zero

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

Divide both sides by

- 1

- u = - 2

Divide both sides by

- 1

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

Simplify

u = 2

u = 2

The following points are undefined

u = 2

Combine undefined points with solutions:

u = 1

Substitute back

u = sin (x)

sin (x) = 1

sin (x) = 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

General solutions for

sin (x) = 1

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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1/((2-sin(x)))=sin(x) ?
The general solution for 1/((2-sin(x)))=sin(x) is x= pi/2+2pin