What is the general solution for 2cot^2(3θ)=5csc(3θ)-4 ?

The general solution for 2cot^2(3θ)=5csc(3θ)-4 is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3

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2cot^2(3θ)=5csc(3θ)-4

Solution

2 cot^{2} (3 θ) = 5 csc (3 θ) - 4

Solution

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

Degrees

θ = 1 0^{\circ} + 12 0^{\circ} n, θ = 5 0^{\circ} + 12 0^{\circ} n

Solution steps

2 cot^{2} (3 θ) = 5 csc (3 θ) - 4

Subtract

5 csc (3 θ) - 4

from both sides

2 cot^{2} (3 θ) - 5 csc (3 θ) + 4 = 0

Rewrite using trig identities

4 + 2 cot^{2} (3 θ) - 5 csc (3 θ)

Use the Pythagorean identity:

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

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

= 4 + 2 (csc^{2} (3 θ) - 1) - 5 csc (3 θ)

Simplify

4 + 2 (csc^{2} (3 θ) - 1) - 5 csc (3 θ) : 2 csc^{2} (3 θ) - 5 csc (3 θ) + 2

4 + 2 (csc^{2} (3 θ) - 1) - 5 csc (3 θ)

Expand

2 (csc^{2} (3 θ) - 1) : 2 csc^{2} (3 θ) - 2

2 (csc^{2} (3 θ) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = csc^{2} (3 θ), c = 1

= 2 csc^{2} (3 θ) - 2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 csc^{2} (3 θ) - 2

= 4 + 2 csc^{2} (3 θ) - 2 - 5 csc (3 θ)

Simplify

4 + 2 csc^{2} (3 θ) - 2 - 5 csc (3 θ) : 2 csc^{2} (3 θ) - 5 csc (3 θ) + 2

4 + 2 csc^{2} (3 θ) - 2 - 5 csc (3 θ)

Group like terms

= 2 csc^{2} (3 θ) - 5 csc (3 θ) + 4 - 2

Add/Subtract the numbers:

4 - 2 = 2

= 2 csc^{2} (3 θ) - 5 csc (3 θ) + 2

= 2 csc^{2} (3 θ) - 5 csc (3 θ) + 2

= 2 csc^{2} (3 θ) - 5 csc (3 θ) + 2

2 + 2 csc^{2} (3 θ) - 5 csc (3 θ) = 0

Solve by substitution

2 + 2 csc^{2} (3 θ) - 5 csc (3 θ) = 0

Let:

csc (3 θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

= 5^{2} - 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

Subtract the numbers:

25 - 16 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

5 + 3 = 8

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

5 - 3 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = csc (3 θ)

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

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

csc (3 θ) = 2 : θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

csc (3 θ) = 2

General solutions for

csc (3 θ) = 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

3 θ = \frac{π}{6} + 2 πn, 3 θ = \frac{5 π}{6} + 2 πn

3 θ = \frac{π}{6} + 2 πn, 3 θ = \frac{5 π}{6} + 2 πn

Solve

3 θ = \frac{π}{6} + 2 πn : θ = \frac{π}{18} + \frac{2 πn}{3}

3 θ = \frac{π}{6} + 2 πn

Divide both sides by

3

3 θ = \frac{π}{6} + 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{π}{6}}{3} + \frac{2 πn}{3} : \frac{π}{18} + \frac{2 πn}{3}

\frac{\frac{π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}

Solve

3 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{18} + \frac{2 πn}{3}

3 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

3

3 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3} : \frac{5 π}{18} + \frac{2 πn}{3}

\frac{\frac{5 π}{6}}{3} + \frac{2 πn}{3}

\frac{\frac{5 π}{6}}{3} = \frac{5 π}{18}

\frac{\frac{5 π}{6}}{3}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{5 π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{5 π}{18}

= \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{5 π}{18} + \frac{2 πn}{3}

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

csc (3 θ) = \frac{1}{2} :

No Solution

csc (3 θ) = \frac{1}{2}

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

No Solution

Combine all the solutions

θ = \frac{π}{18} + \frac{2 πn}{3}, θ = \frac{5 π}{18} + \frac{2 πn}{3}

Popular Examples

1-cos^2(x/2)=cos^2(x)

1 - cos^{2} (\frac{x}{2}) = cos^{2} (x)

1=2sin(2θ)

1 = 2 sin (2 θ)

cos(-x)=(sqrt(3))/2

cos (- x) = \frac{3}{2}

sin(x)=0.61366

sin (x) = 0.61366

sin(α)=-15/17

sin (α) = - \frac{15}{17}

Frequently Asked Questions (FAQ)

What is the general solution for 2cot^2(3θ)=5csc(3θ)-4 ?
The general solution for 2cot^2(3θ)=5csc(3θ)-4 is θ= pi/(18)+(2pin)/3 ,θ=(5pi}{18}+\frac{2pin)/3