What is the general solution for 5cot^2(x)-2csc(x)+2=0 ?

The general solution for 5cot^2(x)-2csc(x)+2=0 is x= pi/2+2pin

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5cot^2(x)-2csc(x)+2=0

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 2 - 2 csc (x) + 5 (csc^{2} (x) - 1)

Simplify

2 - 2 csc (x) + 5 (csc^{2} (x) - 1) : 5 csc^{2} (x) - 2 csc (x) - 3

2 - 2 csc (x) + 5 (csc^{2} (x) - 1)

Expand

5 (csc^{2} (x) - 1) : 5 csc^{2} (x) - 5

5 (csc^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

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

= 5 csc^{2} (x) - 5 \cdot 1

Multiply the numbers:

5 \cdot 1 = 5

= 5 csc^{2} (x) - 5

= 2 - 2 csc (x) + 5 csc^{2} (x) - 5

Simplify

2 - 2 csc (x) + 5 csc^{2} (x) - 5 : 5 csc^{2} (x) - 2 csc (x) - 3

2 - 2 csc (x) + 5 csc^{2} (x) - 5

Group like terms

= - 2 csc (x) + 5 csc^{2} (x) + 2 - 5

Add/Subtract the numbers:

2 - 5 = - 3

= 5 csc^{2} (x) - 2 csc (x) - 3

= 5 csc^{2} (x) - 2 csc (x) - 3

= 5 csc^{2} (x) - 2 csc (x) - 3

- 3 - 2 csc (x) + 5 csc^{2} (x) = 0

Solve by substitution

- 3 - 2 csc (x) + 5 csc^{2} (x) = 0

Let:

csc (x) = u

- 3 - 2 u + 5 u^{2} = 0

- 3 - 2 u + 5 u^{2} = 0 : u = 1, u = - \frac{3}{5}

- 3 - 2 u + 5 u^{2} = 0

Write in the standard form

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

5 u^{2} - 2 u - 3 = 0

Solve with the quadratic formula

5 u^{2} - 2 u - 3 = 0

Quadratic Equation Formula:

For

a = 5, b = - 2, c = - 3

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 2)^{2} - 4 \cdot 5 (- 3) = 8

(- 2)^{2} - 4 \cdot 5 (- 3)

Apply rule

- (- a) = a

= (- 2)^{2} + 4 \cdot 5 \cdot 3

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 5 \cdot 3

Multiply the numbers:

4 \cdot 5 \cdot 3 = 60

= 2^{2} + 60

2^{2} = 4

= 4 + 60

Add the numbers:

4 + 60 = 64

= 64

Factor the number:

64 = 8^{2}

= 8^{2}

Apply radical rule:

8^{2} = 8

= 8

u_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 5}, u_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 5}

u = \frac{- ( - 2 ) + 8}{2 \cdot 5} : 1

\frac{- ( - 2 ) + 8}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{2 + 8}{2 \cdot 5}

Add the numbers:

2 + 8 = 10

= \frac{10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 8}{2 \cdot 5} : - \frac{3}{5}

\frac{- ( - 2 ) - 8}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{2 - 8}{2 \cdot 5}

Subtract the numbers:

2 - 8 = - 6

= \frac{- 6}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{- 6}{10}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{6}{10}

Cancel the common factor:

2

= - \frac{3}{5}

The solutions to the quadratic equation are:

u = 1, u = - \frac{3}{5}

Substitute back

u = csc (x)

csc (x) = 1, csc (x) = - \frac{3}{5}

csc (x) = 1, csc (x) = - \frac{3}{5}

csc (x) = 1 : x = \frac{π}{2} + 2 πn

csc (x) = 1

General solutions for

csc (x) = 1

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

csc (x) = - \frac{3}{5} :

No Solution

csc (x) = - \frac{3}{5}

csc (x) \leq - 1 or csc (x) \geq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5cot^2(x)-2csc(x)+2=0 ?
The general solution for 5cot^2(x)-2csc(x)+2=0 is x= pi/2+2pin