What is the general solution for ((1+cot^2(x)))/(cos^2(x))=cot^2(x) ?

The general solution for ((1+cot^2(x)))/(cos^2(x))=cot^2(x) is No Solution for x\in\mathbb{R}

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Popular Trigonometry >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

cot^{2} (x)

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

= - cos^{2} (x) cot^{2} (x) + csc^{2} (x)

csc^{2} (x) - cos^{2} (x) cot^{2} (x) = 0

Factor

csc^{2} (x) - cos^{2} (x) cot^{2} (x) : (csc (x) + cos (x) cot (x)) (csc (x) - cos (x) cot (x))

csc^{2} (x) - cos^{2} (x) cot^{2} (x)

Rewrite

cos^{2} (x) cot^{2} (x)

(cos (x) cot (x))^{2}

cos^{2} (x) cot^{2} (x)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

= csc^{2} (x) - (cos (x) cot (x))^{2}

Apply Difference of Two Squares Formula:

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

csc^{2} (x) - (cos (x) cot (x))^{2} = (csc (x) + cos (x) cot (x)) (csc (x) - cos (x) cot (x))

= (csc (x) + cos (x) cot (x)) (csc (x) - cos (x) cot (x))

(csc (x) + cos (x) cot (x)) (csc (x) - cos (x) cot (x)) = 0

Solving each part separately

csc (x) + cos (x) cot (x) = 0 or csc (x) - cos (x) cot (x) = 0

csc (x) + cos (x) cot (x) = 0 :

No Solution

csc (x) + cos (x) cot (x) = 0

Express with sin, cos

csc (x) + cos (x) cot (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{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 )}

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

Apply rule

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

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

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

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

1 + u^{2} = 0

Move

1

to the right side

1 + u^{2} = 0

Subtract

1

from 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

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

Substitute back

u = cos (x)

cos (x) = i, cos (x) = - i

cos (x) = i, cos (x) = - i

cos (x) = i :

No Solution

cos (x) = i

No Solution

cos (x) = - i :

No Solution

cos (x) = - i

No Solution

Combine all the solutions

No Solution

csc (x) - cos (x) cot (x) = 0 :

No Solution

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

Express with sin, cos

csc (x) - cos (x) cot (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{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 )}

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

Apply rule

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

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

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

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

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution

Combine all the solutions

No Solution for x \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for ((1+cot^2(x)))/(cos^2(x))=cot^2(x) ?
The general solution for ((1+cot^2(x)))/(cos^2(x))=cot^2(x) is No Solution for x\in\mathbb{R}