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

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

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

Solution

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

Subtract

sec (x)

from both sides

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

Simplify

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

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

Convert element to fraction:

sec (x) = \frac{sec ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

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

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 ) - sec ( x ) ( 1 + sin ( x ) )}{1 + sin ( x )}

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

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

tan (x) + cos (x) - sec (x) (1 + sin (x)) = 0

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

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

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

Multiply fractions:

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

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

1 \cdot (1 + sin (x)) = 1 + sin (x)

1 \cdot (1 + sin (x))

Multiply:

1 \cdot (1 + sin (x)) = (1 + sin (x))

= (1 + sin (x))

Remove parentheses:

(a) = a

= 1 + sin (x)

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

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

Combine the fractions

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

Apply rule

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

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

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

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

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

Expand

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

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

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

- (1 + sin (x))

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

= - 1 - sin (x)

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

Simplify

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

sin (x) - 1 - sin (x)

Group like terms

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

Add similar elements:

sin (x) - sin (x) = 0

= - 1

= - 1

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (x) cos (x) - 1

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^{2} (x) - 1

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

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

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

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

- 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

Popular Examples

sin^2(x)=2+2cos(x),0<= x<= 2pi

sin^{2} (x) = 2 + 2 cos (x), 0 \leq x \leq 2 π

cos(5x+3)= 1/2

cos (5 x + 3) = \frac{1}{2}

csc(θ)=3

csc (θ) = 3

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

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

-3-2cos(θ)=2

- 3 - 2 cos (θ) = 2

Frequently Asked Questions (FAQ)

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