What is the general solution for 2cot^2(y)+3csc(y)=0 ?

The general solution for 2cot^2(y)+3csc(y)=0 is y=(7pi)/6+2pin,y=(11pi)/6+2pin

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2cot^2(y)+3csc(y)=0

Solution

2 cot^{2} (y) + 3 csc (y) = 0

Solution

y = \frac{7 π}{6} + 2 πn, y = \frac{11 π}{6} + 2 πn

Degrees

y = 21 0^{\circ} + 36 0^{\circ} n, y = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cot^{2} (y) + 3 csc (y) = 0

Rewrite using trig identities

2 cot^{2} (y) + 3 csc (y)

Use the Pythagorean identity:

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

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

= 2 (csc^{2} (y) - 1) + 3 csc (y)

(- 1 + csc^{2} (y)) \cdot 2 + 3 csc (y) = 0

Solve by substitution

(- 1 + csc^{2} (y)) \cdot 2 + 3 csc (y) = 0

Let:

csc (y) = u

(- 1 + u^{2}) \cdot 2 + 3 u = 0

(- 1 + u^{2}) \cdot 2 + 3 u = 0 : u = \frac{1}{2}, u = - 2

(- 1 + u^{2}) \cdot 2 + 3 u = 0

Expand

(- 1 + u^{2}) \cdot 2 + 3 u : - 2 + 2 u^{2} + 3 u

(- 1 + u^{2}) \cdot 2 + 3 u

= 2 (- 1 + u^{2}) + 3 u

Expand

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

2 (- 1 + u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = - 1, c = u^{2}

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

Apply minus-plus rules

+ (- a) = - a

= - 2 \cdot 1 + 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 5 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

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

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

Subtract the numbers:

- 3 - 5 = - 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = csc (y)

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

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

csc (y) = \frac{1}{2} :

No Solution

csc (y) = \frac{1}{2}

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

No Solution

csc (y) = - 2 : y = \frac{7 π}{6} + 2 πn, y = \frac{11 π}{6} + 2 πn

csc (y) = - 2

General solutions for

csc (y) = - 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

y = \frac{7 π}{6} + 2 πn, y = \frac{11 π}{6} + 2 πn

y = \frac{7 π}{6} + 2 πn, y = \frac{11 π}{6} + 2 πn

Combine all the solutions

y = \frac{7 π}{6} + 2 πn, y = \frac{11 π}{6} + 2 πn

Popular Examples

cot(x)=-0.456

cot (x) = - 0.456

4tan^2(θ)-2tan(θ)+3=-7tan(θ)+2

4 tan^{2} (θ) - 2 tan (θ) + 3 = - 7 tan (θ) + 2

-2=csc(θ)

- 2 = csc (θ)

2cot(x)=csc^2(x)

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

2csc(2x)cot(2x)=0

2 csc (2 x) cot (2 x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2cot^2(y)+3csc(y)=0 ?
The general solution for 2cot^2(y)+3csc(y)=0 is y=(7pi)/6+2pin,y=(11pi)/6+2pin