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

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

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Popular Trigonometry >

(tan(x)+cos(x))/((1+sin(x)))= 1/(cos(x))

Solution

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

Subtract

\frac{1}{cos ( x )}

from both sides

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

Simplify

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

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

Least Common Multiplier of

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

1 + sin (x), cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + sin (x)

cos (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{tan ( x ) + cos ( x )}{1 + sin ( x )} :

multiply the denominator and numerator by

cos (x)

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

For

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

multiply the denominator and numerator by

sin (x) + 1

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

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

Expand

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

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

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

Expand

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

cos (x) (tan (x) + cos (x))

Apply the distributive law:

a (b + c) = ab + a c

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

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

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

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

- (sin (x) + 1)

Distribute parentheses

= - (sin (x)) - (1)

Apply minus-plus rules

+ (- a) = - a

= - sin (x) - 1

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

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= sin (x)

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

Add similar elements:

- sin (x) + sin (x) = 0

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

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

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

Solve by substitution

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

Let:

cos (x) = u

- 1 + u^{2} = 0

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

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

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

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

Graph

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Frequently Asked Questions (FAQ)

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