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

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

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Popular >

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

Solution

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

Subtract

cos (x)

from both sides

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Combine the fractions

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

Apply rule

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

cos (x) sin (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

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

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

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

Factor

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

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

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

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

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

Factor out common term

sin (x)

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

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

Solving each part separately

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

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

1 + sin (x) - cos^{2} (x) = 0 : x = 2 πn, x = π + 2 πn, x = \frac{3 π}{2} + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

u + u^{2} = 0

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

u + u^{2} = 0

Write in the standard form

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

u^{2} + u = 0

Solve with the quadratic formula

u^{2} + u = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 0

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

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

1^{2} - 4 \cdot 1 \cdot 0 = 1

1^{2} - 4 \cdot 1 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- 1 + 1}{2 \cdot 1} : 0

\frac{- 1 + 1}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

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

= 0

u = \frac{- 1 - 1}{2 \cdot 1} : - 1

\frac{- 1 - 1}{2 \cdot 1}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 0, u = - 1

Substitute back

u = sin (x)

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

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

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

sin (x) = - 1 : x = \frac{3 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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Combine all the solutions

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

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = 2 πn, x = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

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