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

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

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Popular >

1/(tan^2(x))-cos^2(x)=cos^2(x)-cot^2(x)

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

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

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

Cancel the common factor:

sin^{2} (x)

= 1

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Solve by substitution

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

Let:

sin (x) = u

2 - 2 u^{2} = 0

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

2 - 2 u^{2} = 0

Move

2

to the right side

2 - 2 u^{2} = 0

Subtract

2

from both sides

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

Simplify

- 2 u^{2} = - 2

- 2 u^{2} = - 2

Divide both sides by

- 2

- 2 u^{2} = - 2

Divide both sides by

- 2

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

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 = sin (x)

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

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

sin (x) = 1 : x = \frac{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{π}{2} + 2 πn

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

Since the equation is undefined for:

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

No Solution for x \in R

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Frequently Asked Questions (FAQ)

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