What is the general solution for (cos(x)*cot(x))/(1-sin(x))=3 ?

The general solution for (cos(x)*cot(x))/(1-sin(x))=3 is x= pi/6+2pin,x=(5pi)/6+2pin

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Popular Trigonometry >

(cos(x)*cot(x))/(1-sin(x))=3

Solution

\frac{cos ( x ) \cdot cot ( x )}{1 - sin ( x )} = 3

Solution

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

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} = 3

Subtract

3

from both sides

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} - 3 = 0

Simplify

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} - 3 : \frac{cos ( x ) cot ( x ) - 3 ( 1 - sin ( x ) )}{1 - sin ( x )}

\frac{cos ( x ) cot ( x )}{1 - sin ( x )} - 3

Convert element to fraction:

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

= \frac{cos ( x ) cot ( x )}{1 - sin ( x )} - \frac{3 ( 1 - sin ( x ) )}{1 - sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( x ) cot ( x ) - 3 ( 1 - sin ( x ) )}{1 - sin ( x )}

\frac{cos ( x ) cot ( x ) - 3 ( 1 - sin ( x ) )}{1 - sin ( x )} = 0

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

cos (x) cot (x) - 3 (1 - sin (x)) = 0

Rewrite using trig identities

- (1 - sin (x)) \cdot 3 + cos (x) cot (x)

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

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

cos (x) \frac{cos ( x )}{sin ( x )}

Multiply fractions:

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

= \frac{cos ( x ) cos ( x )}{sin ( x )}

cos (x) cos (x) = cos^{2} (x)

cos (x) cos (x)

Apply exponent rule:

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

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Use the Pythagorean identity:

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

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

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

Simplify

\frac{1 - sin ^{2} ( x )}{sin ( x )} - (1 - sin (x)) \cdot 3 : \frac{1 + 2 sin ^{2} ( x )}{sin ( x )} - 3

\frac{1 - sin ^{2} ( x )}{sin ( x )} - (1 - sin (x)) \cdot 3

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

Expand

- 3 (1 - sin (x)) : - 3 + 3 sin (x)

- 3 (1 - sin (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin (x)

= - 3 \cdot 1 - (- 3) sin (x)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 sin (x)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 sin (x)

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

Combine the fractions

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

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

Convert element to fraction:

3 sin (x) = \frac{3 sin ( x ) sin ( x )}{sin ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin^{2} (x) + 3 sin (x) sin (x)

3 sin (x) sin (x) = 3 sin^{2} (x)

3 sin (x) sin (x)

Apply exponent rule:

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

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

= 3 sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 3 sin^{2} (x)

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

Add similar elements:

- sin^{2} (x) + 3 sin^{2} (x) = 2 sin^{2} (x)

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

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

- 3 u + \frac{1 + 2 u ^{2}}{u} u = 0 \cdot u

Simplify

- 3 u + \frac{1 + 2 u ^{2}}{u} u = 0 \cdot u

Simplify

\frac{1 + 2 u ^{2}}{u} u : 1 + 2 u^{2}

\frac{1 + 2 u ^{2}}{u} u

Multiply fractions:

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

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

Cancel the common factor:

u

= 1 + 2 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

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

Write in the standard form

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

2 u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

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

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = sin (x)

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

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

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) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 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{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

Combine all the solutions

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

Since the equation is undefined for:

\frac{π}{2} + 2 πn

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (cos(x)*cot(x))/(1-sin(x))=3 ?
The general solution for (cos(x)*cot(x))/(1-sin(x))=3 is x= pi/6+2pin,x=(5pi)/6+2pin