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

The general solution for (cos^2(x)+1)/(1+cot^2(x))=1 is x= pi/2+2pin,x=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

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

1 - 1 = 0

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

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

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

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

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

Factor

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

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

cos (x) + cot (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) + cot (x) = 0

Express with sin, cos

cos (x) + cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

cos (x) + cos (x) sin (x) = 0

Factor

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

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

Factor out common term

cos (x)

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

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

Solving each part separately

cos (x) = 0 or sin (x) + 1 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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

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

sin (x) + 1 = 0 : x = \frac{3 π}{2} + 2 πn

sin (x) + 1 = 0

Move

1

to the right side

sin (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

sin (x) = - 1

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

cos (x) - cot (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) - cot (x) = 0

Express with sin, cos

cos (x) - cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Factor

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

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

Factor out common term

cos (x)

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

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

Solving each part separately

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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

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

sin (x) - 1 = 0 : x = \frac{π}{2} + 2 πn

sin (x) - 1 = 0

Move

1

to the right side

sin (x) - 1 = 0

Add

1

to both sides

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

Simplify

sin (x) = 1

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

Combine all the solutions

x = \frac{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

Popular Examples

3sin^2(c)-7sin(x)+2=0

3 sin^{2} (c) - 7 sin (x) + 2 = 0

1+cos^2(x)=sin^4(x)

1 + cos^{2} (x) = sin^{4} (x)

cos(t)=cos(2t)

cos (t) = cos (2 t)

arctan(x+2)=arcsin(7/25)+arccos(4/5)

arctan (x + 2) = arcsin (\frac{7}{25}) + arccos (\frac{4}{5})

sec(a)= 38/13

sec (a) = \frac{38}{13}

Frequently Asked Questions (FAQ)

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