What is the general solution for 3csc^2(x)-5csc(x)-2=0 ?

The general solution for 3csc^2(x)-5csc(x)-2=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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3csc^2(x)-5csc(x)-2=0

Solution

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

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

csc (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 5)^{2} - 4 \cdot 3 (- 2) = 7

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 5 ) \pm 7}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 7}{2 \cdot 3}, u_{2} = \frac{- ( - 5 ) - 7}{2 \cdot 3}

u = \frac{- ( - 5 ) + 7}{2 \cdot 3} : 2

\frac{- ( - 5 ) + 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{5 + 7}{2 \cdot 3}

Add the numbers:

5 + 7 = 12

= \frac{12}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{12}{6}

Divide the numbers:

\frac{12}{6} = 2

= 2

u = \frac{- ( - 5 ) - 7}{2 \cdot 3} : - \frac{1}{3}

\frac{- ( - 5 ) - 7}{2 \cdot 3}

Apply rule

- (- a) = a

= \frac{5 - 7}{2 \cdot 3}

Subtract the numbers:

5 - 7 = - 2

= \frac{- 2}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 2}{6}

Apply the fraction rule:

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

= - \frac{2}{6}

Cancel the common factor:

2

= - \frac{1}{3}

The solutions to the quadratic equation are:

u = 2, u = - \frac{1}{3}

Substitute back

u = csc (x)

csc (x) = 2, csc (x) = - \frac{1}{3}

csc (x) = 2, csc (x) = - \frac{1}{3}

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

General solutions for

csc (x) = 2

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

No Solution

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

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

No Solution

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3csc^2(x)-5csc(x)-2=0 ?
The general solution for 3csc^2(x)-5csc(x)-2=0 is x= pi/6+2pin,x=(5pi)/6+2pin