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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

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

\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))) = 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

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

3 - sin (θ) = cos (2 θ)

tan (θ) = \frac{- 1}{1}

cot (x) = 3

cos (\frac{θ}{2}) = - 1

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

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