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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1-sin(x))/(cos(x))=(cos(x))/(1-sin(x))

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

x = 2 πn, x = π + 2 πn

Solution steps

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

Subtract

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

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = 1, sin (x) = 0

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = 2 πn, x = π + 2 πn

5 sec (x) + 7 = - 3

cos (2 θ) + 2 cos (θ) = - 1

(\frac{tan ( x )}{1})^{- 1} = \frac{20}{9.7}

1 - tan (x) = sec (x)

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π

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