What is the general solution for cot(x)cos(x)-sin(x)=1 ?

The general solution for cot(x)cos(x)-sin(x)=1 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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

Solution

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

Solution

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

cot (x) cos (x) - sin (x) - 1 = 0

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

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

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

Use the Pythagorean identity:

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

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

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

Combine the fractions

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Add similar elements:

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

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

Simplify

- 1 \cdot u : - u

- 1 \cdot u

Multiply:

1 \cdot u = u

= - u

Simplify

\frac{1 - 2 u ^{2}}{u} u : 1 - 2 u^{2}

\frac{1 - 2 u ^{2}}{u} u

Multiply fractions:

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

= \frac{( 1 - 2 u ^{2} ) u}{u}

Cancel the common factor:

u

= 1 - 2 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 2) \cdot 1 = 3

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

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \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 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 3}{2 ( - 2 )} : - 1

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 3}{- 2 \cdot 2}

Add the numbers:

1 + 3 = 4

= \frac{4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 3}{- 2 \cdot 2}

Subtract the numbers:

1 - 3 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 1 + \frac{1 - 2 u ^{2}}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = sin (x)

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

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

General solutions for

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

cos^4(x)+cos^3(x)-2=0

cos^{4} (x) + cos^{3} (x) - 2 = 0

tan^2(x)= 1/(cos(x)+1)

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

(sin^2(x)-2cos(x)+1)/4 =0

\frac{sin ^{2} ( x ) - 2 cos ( x ) + 1}{4} = 0

cos^2(x)+cos^4(x)+cos^6(x)=0

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

sin(x)=(-1)/4

sin (x) = \frac{- 1}{4}

Frequently Asked Questions (FAQ)

What is the general solution for cot(x)cos(x)-sin(x)=1 ?
The general solution for cot(x)cos(x)-sin(x)=1 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin