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

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

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(sin(x))/(1+cos(x))+cot(x)=2

Solution

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

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

Subtract

2

from both sides

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

Simplify

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

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

Convert element to fraction:

cot (x) = \frac{cot ( x ) ( 1 + cos ( x ) )}{1 + cos ( x )}, 2 = \frac{2 ( 1 + cos ( x ) )}{1 + cos ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

sin (x) + cot (x) (1 + cos (x)) - 2 (1 + cos (x)) = 0

Express with sin, cos

sin (x) - (1 + cos (x)) \cdot 2 + (1 + cos (x)) cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

(1 + cos (x)) \frac{cos ( x )}{sin ( x )} : \frac{cos ( x ) ( cos ( x ) + 1 )}{sin ( x )}

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

Multiply fractions:

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (x) sin (x) - (1 + cos (x)) \cdot 2 sin (x) + cos (x) (1 + cos (x))

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Expand

cos (x) (1 + cos (x)) : cos (x) + cos^{2} (x)

cos (x) (1 + cos (x))

Apply the distributive law:

a (b + c) = ab + a c

a = cos (x), b = 1, c = cos (x)

= cos (x) \cdot 1 + cos (x) cos (x)

= 1 \cdot cos (x) + cos (x) cos (x)

Simplify

1 \cdot cos (x) + cos (x) cos (x) : cos (x) + cos^{2} (x)

1 \cdot cos (x) + cos (x) cos (x)

1 \cdot cos (x) = cos (x)

1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (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)

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

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

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

Expand

- 2 sin (x) (1 + cos (x)) : - 2 sin (x) - 2 sin (x) cos (x)

- 2 sin (x) (1 + cos (x))

Apply the distributive law:

a (b + c) = ab + a c

a = - 2 sin (x), b = 1, c = cos (x)

= - 2 sin (x) \cdot 1 + (- 2 sin (x)) cos (x)

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 1 \cdot sin (x) - 2 sin (x) cos (x)

Multiply the numbers:

2 \cdot 1 = 2

= - 2 sin (x) - 2 sin (x) cos (x)

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

Simplify

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

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

Group like terms

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

Add similar elements:

- cos^{2} (x) + cos^{2} (x) = 0

= cos (x) - 2 sin (x) - 2 sin (x) cos (x) + 1

= cos (x) - 2 sin (x) - 2 sin (x) cos (x) + 1

= cos (x) - 2 sin (x) - 2 sin (x) cos (x) + 1

1 + cos (x) - 2 sin (x) - 2 cos (x) sin (x) = 0

Factor

1 + cos (x) - 2 sin (x) - 2 cos (x) sin (x) : (1 - 2 sin (x)) (cos (x) + 1)

1 + cos (x) - 2 sin (x) - 2 cos (x) sin (x)

Factor out common term

cos (x)

= 1 + cos (x) (1 - 2 sin (x)) - 2 sin (x)

Rewrite as

= (1 - 2 sin (x)) cos (x) + 1 \cdot (1 - 2 sin (x))

Factor out common term

(1 - 2 sin (x))

= (1 - 2 sin (x)) (cos (x) + 1)

(1 - 2 sin (x)) (cos (x) + 1) = 0

Solving each part separately

1 - 2 sin (x) = 0 or cos (x) + 1 = 0

1 - 2 sin (x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

1 - 2 sin (x) = 0

Move

1

to the right side

1 - 2 sin (x) = 0

Subtract

1

from both sides

1 - 2 sin (x) - 1 = 0 - 1

Simplify

- 2 sin (x) = - 1

- 2 sin (x) = - 1

Divide both sides by

- 2

- 2 sin (x) = - 1

Divide both sides by

- 2

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

Simplify

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

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

cos (x) + 1 = 0 : x = π + 2 πn

cos (x) + 1 = 0

Move

1

to the right side

cos (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

cos (x) = - 1

cos (x) = - 1

General solutions for

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

π + 2 πn

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

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Frequently Asked Questions (FAQ)

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