What is the general solution for 3csc^2(x)+1.5cot(x)=15 ?

The general solution for 3csc^2(x)+1.5cot(x)=15 is x=0.51534…+pin,x=2.72592…+pin

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Popular >

3csc^2(x)+1.5cot(x)=15

Solution

3 csc^{2} (x) + 1.5 cot (x) = 15

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

3 csc^{2} (x) + 1.5 cot (x) = 15

Subtract

15

from both sides

3 csc^{2} (x) + 1.5 cot (x) - 15 = 0

Rewrite using trig identities

- 15 + 1.5 cot (x) + 3 csc^{2} (x)

Use the Pythagorean identity:

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

= - 15 + 1.5 cot (x) + 3 (1 + cot^{2} (x))

Simplify

- 15 + 1.5 cot (x) + 3 (1 + cot^{2} (x)) : 3 cot^{2} (x) + 1.5 cot (x) - 12

- 15 + 1.5 cot (x) + 3 (1 + cot^{2} (x))

Expand

3 (1 + cot^{2} (x)) : 3 + 3 cot^{2} (x)

3 (1 + cot^{2} (x))

Apply the distributive law:

a (b + c) = ab + a c

a = 3, b = 1, c = cot^{2} (x)

= 3 \cdot 1 + 3 cot^{2} (x)

Multiply the numbers:

3 \cdot 1 = 3

= 3 + 3 cot^{2} (x)

= - 15 + 1.5 cot (x) + 3 + 3 cot^{2} (x)

Simplify

- 15 + 1.5 cot (x) + 3 + 3 cot^{2} (x) : 3 cot^{2} (x) + 1.5 cot (x) - 12

- 15 + 1.5 cot (x) + 3 + 3 cot^{2} (x)

Group like terms

= 1.5 cot (x) + 3 cot^{2} (x) - 15 + 3

Add/Subtract the numbers:

- 15 + 3 = - 12

= 3 cot^{2} (x) + 1.5 cot (x) - 12

= 3 cot^{2} (x) + 1.5 cot (x) - 12

= 3 cot^{2} (x) + 1.5 cot (x) - 12

- 12 + 1.5 cot (x) + 3 cot^{2} (x) = 0

Solve by substitution

- 12 + 1.5 cot (x) + 3 cot^{2} (x) = 0

Let:

cot (x) = u

- 12 + 1.5 u + 3 u^{2} = 0

- 12 + 1.5 u + 3 u^{2} = 0 : u = \frac{- 1 + 65}{4}, u = - \frac{1 + 65}{4}

- 12 + 1.5 u + 3 u^{2} = 0

Multiply both sides by

10

- 12 + 1.5 u + 3 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

- 12 \cdot 10 + 1.5 u \cdot 10 + 3 u^{2} \cdot 10 = 0 \cdot 10

Refine

- 120 + 15 u + 30 u^{2} = 0

- 120 + 15 u + 30 u^{2} = 0

Write in the standard form

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

30 u^{2} + 15 u - 120 = 0

Solve with the quadratic formula

30 u^{2} + 15 u - 120 = 0

Quadratic Equation Formula:

For

a = 30, b = 15, c = - 120

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 30 ( - 120 )}{2 \cdot 30}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 30 ( - 120 )}{2 \cdot 30}

1 5^{2} - 4 \cdot 30 (- 120) = 1565

1 5^{2} - 4 \cdot 30 (- 120)

Apply rule

- (- a) = a

= 1 5^{2} + 4 \cdot 30 \cdot 120

Multiply the numbers:

4 \cdot 30 \cdot 120 = 14400

= 1 5^{2} + 14400

1 5^{2} = 225

= 225 + 14400

Add the numbers:

225 + 14400 = 14625

= 14625

Prime factorization of

14625 : 3^{2} \cdot 5^{3} \cdot 13

14625

14625

divides by

314625 = 4875 \cdot 3

= 3 \cdot 4875

4875

divides by

34875 = 1625 \cdot 3

= 3 \cdot 3 \cdot 1625

1625

divides by

51625 = 325 \cdot 5

= 3 \cdot 3 \cdot 5 \cdot 325

325

divides by

5325 = 65 \cdot 5

= 3 \cdot 3 \cdot 5 \cdot 5 \cdot 65

65

divides by

565 = 13 \cdot 5

= 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 13

3, 5, 13

are all prime numbers, therefore no further factorization is possible

= 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 13

= 3^{2} \cdot 5^{3} \cdot 13

= 5^{3} \cdot 3^{2} \cdot 13

Apply exponent rule:

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

= 3^{2} \cdot 5^{2} \cdot 5 \cdot 13

Apply radical rule:

= 3^{2} 5^{2} 5 \cdot 13

Apply radical rule:

3^{2} = 3

= 3 5^{2} 5 \cdot 13

Apply radical rule:

5^{2} = 5

= 3 \cdot 5 5 \cdot 13

Refine

= 1565

u_{1, 2} = \frac{- 15 \pm 15 65}{2 \cdot 30}

Separate the solutions

u_{1} = \frac{- 15 + 15 65}{2 \cdot 30}, u_{2} = \frac{- 15 - 15 65}{2 \cdot 30}

u = \frac{- 15 + 15 65}{2 \cdot 30} : \frac{- 1 + 65}{4}

\frac{- 15 + 15 65}{2 \cdot 30}

Multiply the numbers:

2 \cdot 30 = 60

= \frac{- 15 + 15 65}{60}

Factor

- 15 + 1565 : 15 (- 1 + 65)

- 15 + 1565

Rewrite as

= - 15 \cdot 1 + 1565

Factor out common term

15

= 15 (- 1 + 65)

= \frac{15 ( - 1 + 65 )}{60}

Cancel the common factor:

15

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

u = \frac{- 15 - 15 65}{2 \cdot 30} : - \frac{1 + 65}{4}

\frac{- 15 - 15 65}{2 \cdot 30}

Multiply the numbers:

2 \cdot 30 = 60

= \frac{- 15 - 15 65}{60}

Factor

- 15 - 1565 : - 15 (1 + 65)

- 15 - 1565

Rewrite as

= - 15 \cdot 1 - 1565

Factor out common term

15

= - 15 (1 + 65)

= - \frac{15 ( 1 + 65 )}{60}

Cancel the common factor:

15

= - \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 3csc^2(x)+1.5cot(x)=15 ?
The general solution for 3csc^2(x)+1.5cot(x)=15 is x=0.51534…+pin,x=2.72592…+pin