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

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

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1/(cos(x)+1)+1/(cos(x)-1)=2csc(x)cot(x)

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

2 csc (x) cot (x)

from both sides

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Factor

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

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

Factor out common term

2 cos (x)

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

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

Solving each part separately

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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

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

sin^{2} (x) - (- 1 + cos (x)) (1 + cos (x)) = 0 : x = 2 πn, x = π + 2 πn

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

Rewrite using trig identities

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

Simplify

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

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

Expand

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

Expand

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

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

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = cos (x), b = 1

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

Apply rule

1^{a} = 1

1^{2} = 1

= cos^{2} (x) - 1

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Simplify

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

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

Add similar elements:

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

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

Group like terms

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

2 - 2 u^{2} = 0

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

2 - 2 u^{2} = 0

Move

2

to the right side

2 - 2 u^{2} = 0

Subtract

2

from both sides

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

Simplify

- 2 u^{2} = - 2

- 2 u^{2} = - 2

Divide both sides by

- 2

- 2 u^{2} = - 2

Divide both sides by

- 2

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

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 = cos (x)

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

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 = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

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 = 2 πn, x = π + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

2 πn, π + 2 πn

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

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tan (θ) = \frac{- 1}{1}

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cot (x) = 3

cos(θ/2)=-1

cos (\frac{θ}{2}) = - 1

2sin(x)+(sqrt(3))/4 = 3/2 sin(x)

2 sin (x) + \frac{3}{4} = \frac{3}{2} sin (x)

Frequently Asked Questions (FAQ)

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