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

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

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Popular Trigonometry >

cos(x)= 1/(cot(x))

Solution

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

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

Subtract

\frac{1}{cot ( x )}

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Rewrite using trig identities

- 1 + cos (x) cot (x)

Use the basic trigonometric identity:

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

= - 1 + 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 + \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 )}

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

Simplify

- 1 \cdot u : - u

- 1 \cdot u

Multiply:

1 \cdot u = u

= - u

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

u

= 1 - u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

- u + 1 - 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}

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Show solutions in decimal form

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

Graph

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Frequently Asked Questions (FAQ)

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