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

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

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

Least Common Multiplier of

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

cos (x), 1 - sin (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

1 - sin (x)

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

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

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

For

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

multiply the denominator and numerator by

- sin (x) + 1

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

For

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

multiply the denominator and numerator by

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

= \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

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

Apply Perfect Square Formula:

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

a = 1, b = sin (x)

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

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

1 - 1 = 0

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Write in the standard form

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

2 u^{2} - 2 u = 0

Solve with the quadratic formula

2 u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 2 \cdot 0

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

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

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 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{- ( - 2 ) - 2}{2 \cdot 2} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = sin (x)

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

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

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) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, 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

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5 sec (x) + 7 = - 3

cos(2θ)+2cos(θ)=-1

cos (2 θ) + 2 cos (θ) = - 1

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

1-tan(x)=sec(x)

1 - tan (x) = sec (x)

sec((3θ)/2)=-2,0<= θ<2pi

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Frequently Asked Questions (FAQ)

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