What is the general solution for sqrt(1+cot^2(x))=8 ?

The general solution for sqrt(1+cot^2(x))=8 is x=0.12532…+pin,x=3.01626…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(1+cot^2(x))=8

Solution

1 + cot^{2} (x) = 8

Solution

x = 0.12532 \dots + πn, x = 3.01626 \dots + πn

Degrees

x = 7.18075 \dots^{\circ} + 18 0^{\circ} n, x = 172.81924 \dots^{\circ} + 18 0^{\circ} n

Solution steps

1 + cot^{2} (x) = 8

Solve by substitution

1 + cot^{2} (x) = 8

Let:

cot (x) = u

1 + u^{2} = 8

1 + u^{2} = 8 : u = 37, u = - 37

1 + u^{2} = 8

Square both sides:

1 + u^{2} = 64

1 + u^{2} = 8

(1 + u^{2})^{2} = 8^{2}

Expand

(1 + u^{2})^{2} : 1 + u^{2}

(1 + u^{2})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

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

= (1 + u^{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 + u^{2}

Expand

8^{2} : 64

8^{2}

8^{2} = 64

= 64

1 + u^{2} = 64

1 + u^{2} = 64

Solve

1 + u^{2} = 64 : u = 37, u = - 37

1 + u^{2} = 64

Move

1

to the right side

1 + u^{2} = 64

Subtract

1

from both sides

1 + u^{2} - 1 = 64 - 1

Simplify

u^{2} = 63

u^{2} = 63

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 63, u = - 63

63 = 37

63

Prime factorization of

63 : 3^{2} \cdot 7

63

63

divides by

363 = 21 \cdot 3

= 3 \cdot 21

21

divides by

321 = 7 \cdot 3

= 3 \cdot 3 \cdot 7

3, 7

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 3 \cdot 7

= 3^{2} \cdot 7

= 3^{2} \cdot 7

Apply radical rule:

= 7 3^{2}

Apply radical rule:

3^{2} = 3

= 37

- 63 = - 37

- 63

63 = 37

63

Prime factorization of

63 : 3^{2} \cdot 7

63

63

divides by

363 = 21 \cdot 3

= 3 \cdot 21

21

divides by

321 = 7 \cdot 3

= 3 \cdot 3 \cdot 7

3, 7

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 3 \cdot 7

= 3^{2} \cdot 7

= 3^{2} \cdot 7

Apply radical rule:

= 7 3^{2}

Apply radical rule:

3^{2} = 3

= 37

= - 37

u = 37, u = - 37

u = 37, u = - 37

Verify Solutions:

u = 37

True

, u = - 37

True

Check the solutions by plugging them into

1 + u^{2} = 8

Remove the ones that don't agree with the equation.

Plug in

u = 37 :

True

1 + (37)^{2} = 8

1 + (37)^{2} = 8

1 + (37)^{2}

(37)^{2} = 3^{2} \cdot 7

(37)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 3^{2} (7)^{2}

(7)^{2} : 7

Apply radical rule:

a = a^{\frac{1}{2}}

= (7^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 7^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 7

= 3^{2} \cdot 7

= 1 + 3^{2} \cdot 7

3^{2} \cdot 7 = 63

3^{2} \cdot 7

3^{2} = 9

= 9 \cdot 7

Multiply the numbers:

9 \cdot 7 = 63

= 63

= 1 + 63

Add the numbers:

1 + 63 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

8 = 8

True

Plug in

u = - 37 :

True

1 + (- 37)^{2} = 8

1 + (- 37)^{2} = 8

1 + (- 37)^{2}

(- 37)^{2} = 3^{2} \cdot 7

(- 37)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 37)^{2} = (37)^{2}

= (37)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 3^{2} (7)^{2}

(7)^{2} : 7

Apply radical rule:

a = a^{\frac{1}{2}}

= (7^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 7^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 7

= 3^{2} \cdot 7

= 1 + 3^{2} \cdot 7

3^{2} \cdot 7 = 63

3^{2} \cdot 7

3^{2} = 9

= 9 \cdot 7

Multiply the numbers:

9 \cdot 7 = 63

= 63

= 1 + 63

Add the numbers:

1 + 63 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

8 = 8

True

The solutions are

u = 37, u = - 37

Substitute back

u = cot (x)

cot (x) = 37, cot (x) = - 37

cot (x) = 37, cot (x) = - 37

cot (x) = 37 : x = arccot (37) + πn

cot (x) = 37

Apply trig inverse properties

cot (x) = 37

General solutions for

cot (x) = 37

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (37) + πn

x = arccot (37) + πn

cot (x) = - 37 : x = arccot (- 37) + πn

cot (x) = - 37

Apply trig inverse properties

cot (x) = - 37

General solutions for

cot (x) = - 37

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (- 37) + πn

x = arccot (- 37) + πn

Combine all the solutions

x = arccot (37) + πn, x = arccot (- 37) + πn

Show solutions in decimal form

x = 0.12532 \dots + πn, x = 3.01626 \dots + πn

Popular Examples

1/(1-sin(x))+1/(1+sin(x))=2csc^2(x)

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

tan(x)-sqrt(1-2tan^2(x))=0

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

cos(θ)=-sqrt(2/5),tan(2θ)

cos (θ) = - \frac{2}{5}, tan (2 θ)

sin(x+pi/6)=(sqrt(2))/2

sin (x + \frac{π}{6}) = \frac{2}{2}

1/(sqrt(2))=cos(x)

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

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(1+cot^2(x))=8 ?
The general solution for sqrt(1+cot^2(x))=8 is x=0.12532…+pin,x=3.01626…+pin