What is the general solution for 5csc^2(x)+2.5cot(x)=25 ?

The general solution for 5csc^2(x)+2.5cot(x)=25 is x=0.51534…+pin,x=2.72592…+pin

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5csc^2(x)+2.5cot(x)=25

Solution

5 csc^{2} (x) + 2.5 cot (x) = 25

Solution

x = 0.51534 \dots + πn, x = 2.72592 \dots + πn

Degrees

x = 29.52683 \dots^{\circ} + 18 0^{\circ} n, x = 156.18376 \dots^{\circ} + 18 0^{\circ} n

Solution steps

5 csc^{2} (x) + 2.5 cot (x) = 25

Subtract

25

from both sides

5 csc^{2} (x) + 2.5 cot (x) - 25 = 0

Rewrite using trig identities

- 25 + 2.5 cot (x) + 5 csc^{2} (x)

Use the Pythagorean identity:

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

= - 25 + 2.5 cot (x) + 5 (1 + cot^{2} (x))

Simplify

- 25 + 2.5 cot (x) + 5 (1 + cot^{2} (x)) : 5 cot^{2} (x) + 2.5 cot (x) - 20

- 25 + 2.5 cot (x) + 5 (1 + cot^{2} (x))

Expand

5 (1 + cot^{2} (x)) : 5 + 5 cot^{2} (x)

5 (1 + cot^{2} (x))

Apply the distributive law:

a (b + c) = ab + a c

a = 5, b = 1, c = cot^{2} (x)

= 5 \cdot 1 + 5 cot^{2} (x)

Multiply the numbers:

5 \cdot 1 = 5

= 5 + 5 cot^{2} (x)

= - 25 + 2.5 cot (x) + 5 + 5 cot^{2} (x)

Simplify

- 25 + 2.5 cot (x) + 5 + 5 cot^{2} (x) : 5 cot^{2} (x) + 2.5 cot (x) - 20

- 25 + 2.5 cot (x) + 5 + 5 cot^{2} (x)

Group like terms

= 2.5 cot (x) + 5 cot^{2} (x) - 25 + 5

Add/Subtract the numbers:

- 25 + 5 = - 20

= 5 cot^{2} (x) + 2.5 cot (x) - 20

= 5 cot^{2} (x) + 2.5 cot (x) - 20

= 5 cot^{2} (x) + 2.5 cot (x) - 20

- 20 + 2.5 cot (x) + 5 cot^{2} (x) = 0

Solve by substitution

- 20 + 2.5 cot (x) + 5 cot^{2} (x) = 0

Let:

cot (x) = u

- 20 + 2.5 u + 5 u^{2} = 0

- 20 + 2.5 u + 5 u^{2} = 0 : u = \frac{- 1 + 65}{4}, u = - \frac{1 + 65}{4}

- 20 + 2.5 u + 5 u^{2} = 0

Multiply both sides by

10

- 20 + 2.5 u + 5 u^{2} = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

- 20 \cdot 10 + 2.5 u \cdot 10 + 5 u^{2} \cdot 10 = 0 \cdot 10

Refine

- 200 + 25 u + 50 u^{2} = 0

- 200 + 25 u + 50 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

50 u^{2} + 25 u - 200 = 0

Solve with the quadratic formula

50 u^{2} + 25 u - 200 = 0

Quadratic Equation Formula:

For

a = 50, b = 25, c = - 200

u_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 \cdot 50 ( - 200 )}{2 \cdot 50}

u_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 \cdot 50 ( - 200 )}{2 \cdot 50}

2 5^{2} - 4 \cdot 50 (- 200) = 2565

2 5^{2} - 4 \cdot 50 (- 200)

Apply rule

- (- a) = a

= 2 5^{2} + 4 \cdot 50 \cdot 200

Multiply the numbers:

4 \cdot 50 \cdot 200 = 40000

= 2 5^{2} + 40000

2 5^{2} = 625

= 625 + 40000

Add the numbers:

625 + 40000 = 40625

= 40625

Prime factorization of

40625 : 5^{5} \cdot 13

40625

40625

divides by

540625 = 8125 \cdot 5

= 5 \cdot 8125

8125

divides by

58125 = 1625 \cdot 5

= 5 \cdot 5 \cdot 1625

1625

divides by

51625 = 325 \cdot 5

= 5 \cdot 5 \cdot 5 \cdot 325

325

divides by

5325 = 65 \cdot 5

= 5 \cdot 5 \cdot 5 \cdot 5 \cdot 65

65

divides by

565 = 13 \cdot 5

= 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 13

5, 13

are all prime numbers, therefore no further factorization is possible

= 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 13

= 5^{5} \cdot 13

= 5^{5} \cdot 13

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

= 5^{4} \cdot 5 \cdot 13

Apply radical rule:

= 5^{4} 5 \cdot 13

Apply radical rule:

5^{4} = 5^{\frac{4}{2}} = 5^{2}

= 5^{2} 5 \cdot 13

Refine

= 2565

u_{1, 2} = \frac{- 25 \pm 25 65}{2 \cdot 50}

Separate the solutions

u_{1} = \frac{- 25 + 25 65}{2 \cdot 50}, u_{2} = \frac{- 25 - 25 65}{2 \cdot 50}

u = \frac{- 25 + 25 65}{2 \cdot 50} : \frac{- 1 + 65}{4}

\frac{- 25 + 25 65}{2 \cdot 50}

Multiply the numbers:

2 \cdot 50 = 100

= \frac{- 25 + 25 65}{100}

Factor

- 25 + 2565 : 25 (- 1 + 65)

- 25 + 2565

Rewrite as

= - 25 \cdot 1 + 2565

Factor out common term

25

= 25 (- 1 + 65)

= \frac{25 ( - 1 + 65 )}{100}

Cancel the common factor:

25

= \frac{- 1 + 65}{4}

u = \frac{- 25 - 25 65}{2 \cdot 50} : - \frac{1 + 65}{4}

\frac{- 25 - 25 65}{2 \cdot 50}

Multiply the numbers:

2 \cdot 50 = 100

= \frac{- 25 - 25 65}{100}

Factor

- 25 - 2565 : - 25 (1 + 65)

- 25 - 2565

Rewrite as

= - 25 \cdot 1 - 2565

Factor out common term

25

= - 25 (1 + 65)

= - \frac{25 ( 1 + 65 )}{100}

Cancel the common factor:

25

= - \frac{1 + 65}{4}

The solutions to the quadratic equation are:

u = \frac{- 1 + 65}{4}, u = - \frac{1 + 65}{4}

Substitute back

u = cot (x)

cot (x) = \frac{- 1 + 65}{4}, cot (x) = - \frac{1 + 65}{4}

cot (x) = \frac{- 1 + 65}{4}, cot (x) = - \frac{1 + 65}{4}

cot (x) = \frac{- 1 + 65}{4} : x = arccot (\frac{- 1 + 65}{4}) + πn

cot (x) = \frac{- 1 + 65}{4}

Apply trig inverse properties

cot (x) = \frac{- 1 + 65}{4}

General solutions for

cot (x) = \frac{- 1 + 65}{4}

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

x = arccot (\frac{- 1 + 65}{4}) + πn

x = arccot (\frac{- 1 + 65}{4}) + πn

cot (x) = - \frac{1 + 65}{4} : x = arccot (- \frac{1 + 65}{4}) + πn

cot (x) = - \frac{1 + 65}{4}

Apply trig inverse properties

cot (x) = - \frac{1 + 65}{4}

General solutions for

cot (x) = - \frac{1 + 65}{4}

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

x = arccot (- \frac{1 + 65}{4}) + πn

x = arccot (- \frac{1 + 65}{4}) + πn

Combine all the solutions

x = arccot (\frac{- 1 + 65}{4}) + πn, x = arccot (- \frac{1 + 65}{4}) + πn

Show solutions in decimal form

x = 0.51534 \dots + πn, x = 2.72592 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5csc^2(x)+2.5cot(x)=25 ?
The general solution for 5csc^2(x)+2.5cot(x)=25 is x=0.51534…+pin,x=2.72592…+pin