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

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

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(cot^2(x))/(1+sin(x))=(csc(x)-1)/(sin^2(x))

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

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

Subtract

\frac{csc ( x ) - 1}{sin ^{2} ( x )}

from both sides

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

Simplify

\frac{cot ^{2} ( x )}{1 + sin ( x )} - \frac{csc ( x ) - 1}{sin ^{2} ( x )} : \frac{cot ^{2} ( x ) sin ^{2} ( x ) - ( csc ( x ) - 1 ) ( sin ( x ) + 1 )}{sin ^{2} ( x ) ( sin ( x ) + 1 )}

\frac{cot ^{2} ( x )}{1 + sin ( x )} - \frac{csc ( x ) - 1}{sin ^{2} ( x )}

Least Common Multiplier of

1 + sin (x), sin^{2} (x) : sin^{2} (x) (sin (x) + 1)

1 + sin (x), sin^{2} (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + sin (x)

sin^{2} (x)

= sin^{2} (x) (sin (x) + 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin^{2} (x) (sin (x) + 1)

For

\frac{cot ^{2} ( x )}{1 + sin ( x )} :

multiply the denominator and numerator by

sin^{2} (x)

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

For

\frac{csc ( x ) - 1}{sin ^{2} ( x )} :

multiply the denominator and numerator by

sin (x) + 1

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

cot^{2} (x) sin^{2} (x) - (csc (x) - 1) (sin (x) + 1) = 0

Rewrite using trig identities

- (- 1 + csc (x)) (1 + sin (x)) + cot^{2} (x) sin^{2} (x)

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

cot^{2} (x) = csc^{2} (x) - 1

= - (- 1 + csc (x)) (1 + sin (x)) + (csc^{2} (x) - 1) sin^{2} (x)

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

Factor

- (- 1 + csc (x)) (1 + sin (x)) + (- 1 + csc^{2} (x)) sin^{2} (x) : (- 1 + csc (x)) (- 1 - sin (x) + sin^{2} (x) + sin^{2} (x) csc (x))

- (- 1 + csc (x)) (1 + sin (x)) + (- 1 + csc^{2} (x)) sin^{2} (x)

Factor

- 1 + csc^{2} (x) : (csc (x) + 1) (csc (x) - 1)

- 1 + csc^{2} (x)

Rewrite

1

1^{2}

= csc^{2} (x) - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

csc^{2} (x) - 1^{2} = (csc (x) + 1) (csc (x) - 1)

= (csc (x) + 1) (csc (x) - 1)

= - (csc (x) - 1) (sin (x) + 1) + sin^{2} (x) (csc (x) + 1) (csc (x) - 1)

Factor out common term

(- 1 + csc (x))

= (- 1 + csc (x)) (- (1 + sin (x)) + (1 + csc (x)) sin^{2} (x))

Expand

sin^{2} (x) (csc (x) + 1) - (sin (x) + 1) : - 1 - sin (x) + sin^{2} (x) + sin^{2} (x) csc (x)

- (1 + sin (x)) + (1 + csc (x)) sin^{2} (x)

= - (1 + sin (x)) + sin^{2} (x) (1 + csc (x))

- (1 + sin (x)) : - 1 - sin (x)

- (1 + sin (x))

Distribute parentheses

= - (1) - (sin (x))

Apply minus-plus rules

+ (- a) = - a

= - 1 - sin (x)

= - 1 - sin (x) + (1 + csc (x)) sin^{2} (x)

Expand

sin^{2} (x) (1 + csc (x)) : sin^{2} (x) + sin^{2} (x) csc (x)

sin^{2} (x) (1 + csc (x))

Apply the distributive law:

a (b + c) = ab + a c

a = sin^{2} (x), b = 1, c = csc (x)

= sin^{2} (x) \cdot 1 + sin^{2} (x) csc (x)

= 1 \cdot sin^{2} (x) + sin^{2} (x) csc (x)

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

= sin^{2} (x) + sin^{2} (x) csc (x)

= - 1 - sin (x) + sin^{2} (x) + sin^{2} (x) csc (x)

= (csc (x) - 1) (sin^{2} (x) + sin^{2} (x) csc (x) - sin (x) - 1)

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

Solving each part separately

- 1 + csc (x) = 0 or - 1 - sin (x) + sin^{2} (x) + sin^{2} (x) csc (x) = 0

- 1 + csc (x) = 0 : x = \frac{π}{2} + 2 πn

- 1 + csc (x) = 0

Move

1

to the right side

- 1 + csc (x) = 0

Add

1

to both sides

- 1 + csc (x) + 1 = 0 + 1

Simplify

csc (x) = 1

csc (x) = 1

General solutions for

csc (x) = 1

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

- 1 - sin (x) + sin^{2} (x) + sin^{2} (x) csc (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Rewrite using trig identities

- 1 - sin (x) + sin^{2} (x) + csc (x) sin^{2} (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

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

\frac{1}{sin ( x )} sin^{2} (x)

Multiply fractions:

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

= \frac{1 \cdot sin ^{2} ( x )}{sin ( x )}

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

= \frac{sin ^{2} ( x )}{sin ( x )}

Cancel the common factor:

sin (x)

= sin (x)

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

Group like terms

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

Add similar elements:

- sin (x) + sin (x) = 0

= sin^{2} (x) - 1

= sin^{2} (x) - 1

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

Solve by substitution

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

Let:

sin (x) = u

- 1 + u^{2} = 0

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

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 1

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

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

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

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