What is the general solution for 2/(1+sin(x))=3-2sin(x) ?

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

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2/(1+sin(x))=3-2sin(x)

Solution

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

Solution

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

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Multiply both sides by

1 + u

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

Multiply both sides by

1 + u

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

Simplify

2 = 3 (1 + u) - 2 u (1 + u)

2 = 3 (1 + u) - 2 u (1 + u)

Solve

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

2 = 3 (1 + u) - 2 u (1 + u)

Expand

3 (1 + u) - 2 u (1 + u) : 3 + u - 2 u^{2}

3 (1 + u) - 2 u (1 + u)

Expand

3 (1 + u) : 3 + 3 u

3 (1 + u)

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = 1, c = u

= 3 \cdot 1 + 3 u

Multiply the numbers:

3 \cdot 1 = 3

= 3 + 3 u

= 3 + 3 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 u) u

Apply minus-plus rules

+ (- a) = - a

= - 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}

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

Add similar elements:

3 u - 2 u = u

= 3 + u - 2 u^{2}

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

Switch sides

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

Move

2

to the left side

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

Subtract

2

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 (- 2) \cdot 1 = 3

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

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

\frac{- 1 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \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 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

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

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

Substitute back

u = sin (x)

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

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

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

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

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Frequently Asked Questions (FAQ)

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