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

The general solution for (sin(x)+cos(x))/(cos(x)+1)=tan(x) is x=0.66623…+2pin,x=pi-0.66623…+2pin

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

Solution

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

Solution

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

Degrees

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

tan (x)

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Express with sin, cos

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Convert element to fraction:

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

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

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

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

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

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

Expand

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

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

Expand

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = - sin (x), b = cos (x), c = 1

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

Apply minus-plus rules

+ (- a) = - a

= - sin (x) cos (x) - 1 \cdot sin (x)

Multiply:

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

= - sin (x) cos (x) - sin (x)

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

Add similar elements:

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

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

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

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

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

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

Add

sin (x)

to both sides

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

Square both sides

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

Subtract

sin^{2} (x)

from both sides

cos^{4} (x) - sin^{2} (x) = 0

Factor

cos^{4} (x) - sin^{2} (x) : (cos^{2} (x) + sin (x)) (cos^{2} (x) - sin (x))

cos^{4} (x) - sin^{2} (x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

cos^{4} (x) = (cos^{2} (x))^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

Solving each part separately

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

cos^{2} (x) + sin (x) = 0 : x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 1

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

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

1^{2} - 4 (- 1) \cdot 1 = 5

1^{2} - 4 (- 1) \cdot 1

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 ( - 1 )}, u_{2} = \frac{- 1 - 5}{2 ( - 1 )}

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

The solutions to the quadratic equation are:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}

sin (x) = - \frac{- 1 + 5}{2}, sin (x) = \frac{1 + 5}{2}

sin (x) = - \frac{- 1 + 5}{2} : x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = - \frac{- 1 + 5}{2}

Apply trig inverse properties

sin (x) = - \frac{- 1 + 5}{2}

General solutions for

sin (x) = - \frac{- 1 + 5}{2}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = \frac{1 + 5}{2} :

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

cos^{2} (x) - sin (x) = 0 : x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 - u - u^{2} = 0

1 - u - u^{2} = 0 : u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

1 - u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 1) \cdot 1 = 5

(- 1)^{2} - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 + 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

Apply the fraction rule:

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

= - \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

Remove parentheses:

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

= \frac{1 - 5}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

Apply the fraction rule:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

The solutions to the quadratic equation are:

u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2}, sin (x) = \frac{5 - 1}{2}

sin (x) = - \frac{1 + 5}{2} :

No Solution

sin (x) = - \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = \frac{5 - 1}{2} : x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

sin (x) = \frac{5 - 1}{2}

Apply trig inverse properties

sin (x) = \frac{5 - 1}{2}

General solutions for

sin (x) = \frac{5 - 1}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

Combine all the solutions

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

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

Remove the ones that don't agree with the equation.

Check the solution

arcsin (- \frac{- 1 + 5}{2}) + 2 πn :

False

arcsin (- \frac{- 1 + 5}{2}) + 2 πn

Plug in

n = 1

arcsin (- \frac{- 1 + 5}{2}) + 2 π 1

For

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

plug in

x = arcsin (- \frac{- 1 + 5}{2}) + 2 π 1

\frac{sin ( arcsin ( - \frac{- 1 + 5}{2} ) + 2 π 1 ) + cos ( arcsin ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{cos ( arcsin ( - \frac{- 1 + 5}{2} ) + 2 π 1 ) + 1} = tan (arcsin (- \frac{- 1 + 5}{2}) + 2 π 1)

Refine

0.09412 \dots = - 0.78615 \dots

\Rightarrow False

Check the solution

π + arcsin (\frac{- 1 + 5}{2}) + 2 πn :

False

π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

Plug in

n = 1

π + arcsin (\frac{- 1 + 5}{2}) + 2 π 1

For

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

plug in

x = π + arcsin (\frac{- 1 + 5}{2}) + 2 π 1

\frac{sin ( π + arcsin ( \frac{- 1 + 5}{2} ) + 2 π 1 ) + cos ( π + arcsin ( \frac{- 1 + 5}{2} ) + 2 π 1 )}{cos ( π + arcsin ( \frac{- 1 + 5}{2} ) + 2 π 1 ) + 1} = tan (π + arcsin (\frac{- 1 + 5}{2}) + 2 π 1)

Refine

- 6.56625 \dots = 0.78615 \dots

\Rightarrow False

Check the solution

arcsin (\frac{5 - 1}{2}) + 2 πn :

True

arcsin (\frac{5 - 1}{2}) + 2 πn

Plug in

n = 1

arcsin (\frac{5 - 1}{2}) + 2 π 1

For

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

plug in

x = arcsin (\frac{5 - 1}{2}) + 2 π 1

\frac{sin ( arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + cos ( arcsin ( \frac{5 - 1}{2} ) + 2 π 1 )}{cos ( arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + 1} = tan (arcsin (\frac{5 - 1}{2}) + 2 π 1)

Refine

0.78615 \dots = 0.78615 \dots

\Rightarrow True

Check the solution

π - arcsin (\frac{5 - 1}{2}) + 2 πn :

True

π - arcsin (\frac{5 - 1}{2}) + 2 πn

Plug in

n = 1

π - arcsin (\frac{5 - 1}{2}) + 2 π 1

For

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

plug in

x = π - arcsin (\frac{5 - 1}{2}) + 2 π 1

\frac{sin ( π - arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + cos ( π - arcsin ( \frac{5 - 1}{2} ) + 2 π 1 )}{cos ( π - arcsin ( \frac{5 - 1}{2} ) + 2 π 1 ) + 1} = tan (π - arcsin (\frac{5 - 1}{2}) + 2 π 1)

Refine

- 0.78615 \dots = - 0.78615 \dots

\Rightarrow True

x = arcsin (\frac{5 - 1}{2}) + 2 πn, x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

Show solutions in decimal form

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(x)+cos(x))/(cos(x)+1)=tan(x) ?
The general solution for (sin(x)+cos(x))/(cos(x)+1)=tan(x) is x=0.66623…+2pin,x=pi-0.66623…+2pin