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

The general solution for 2csc^2(θ)-3cot(θ)+3=0 is No Solution for θ\in\mathbb{R}

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Popular >

2csc^2(θ)-3cot(θ)+3=0

Solution

2 csc^{2} (θ) - 3 cot (θ) + 3 = 0

Solution

No Solution for θ \in R

Solution steps

2 csc^{2} (θ) - 3 cot (θ) + 3 = 0

Rewrite using trig identities

3 + 2 csc^{2} (θ) - 3 cot (θ)

Use the Pythagorean identity:

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

= 3 + 2 (1 + cot^{2} (θ)) - 3 cot (θ)

Simplify

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

3 + 2 (1 + cot^{2} (θ)) - 3 cot (θ)

Expand

2 (1 + cot^{2} (θ)) : 2 + 2 cot^{2} (θ)

2 (1 + cot^{2} (θ))

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = 1, c = cot^{2} (θ)

= 2 \cdot 1 + 2 cot^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 + 2 cot^{2} (θ)

= 3 + 2 + 2 cot^{2} (θ) - 3 cot (θ)

Add the numbers:

3 + 2 = 5

= 2 cot^{2} (θ) - 3 cot (θ) + 5

= 2 cot^{2} (θ) - 3 cot (θ) + 5

5 + 2 cot^{2} (θ) - 3 cot (θ) = 0

Solve by substitution

5 + 2 cot^{2} (θ) - 3 cot (θ) = 0

Let:

cot (θ) = u

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

5 + 2 u^{2} - 3 u = 0 : u = \frac{3}{4} + i \frac{31}{4}, u = \frac{3}{4} - i \frac{31}{4}

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 5 : 31 i

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

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

Multiply the numbers:

4 \cdot 2 \cdot 5 = 40

= 3^{2} - 40

Apply imaginary number rule:

- a = i a

= i 40 - 3^{2}

- 3^{2} + 40 = 31

- 3^{2} + 40

3^{2} = 9

= - 9 + 40

Add/Subtract the numbers:

- 9 + 40 = 31

= 31

= 31 i

u_{1, 2} = \frac{- ( - 3 ) \pm 31 i}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 31 i}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 31 i}{2 \cdot 2}

u = \frac{- ( - 3 ) + 31 i}{2 \cdot 2} : \frac{3}{4} + i \frac{31}{4}

\frac{- ( - 3 ) + 31 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 31 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 + 31 i}{4}

Rewrite

\frac{3 + 31 i}{4}

in standard complex form:

\frac{3}{4} + \frac{31}{4} i

\frac{3 + 31 i}{4}

Apply the fraction rule:

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

\frac{3 + 31 i}{4} = \frac{3}{4} + \frac{31 i}{4}

= \frac{3}{4} + \frac{31 i}{4}

= \frac{3}{4} + \frac{31}{4} i

u = \frac{- ( - 3 ) - 31 i}{2 \cdot 2} : \frac{3}{4} - i \frac{31}{4}

\frac{- ( - 3 ) - 31 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 - 31 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 - 31 i}{4}

Rewrite

\frac{3 - 31 i}{4}

in standard complex form:

\frac{3}{4} - \frac{31}{4} i

\frac{3 - 31 i}{4}

Apply the fraction rule:

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

\frac{3 - 31 i}{4} = \frac{3}{4} - \frac{31 i}{4}

= \frac{3}{4} - \frac{31 i}{4}

= \frac{3}{4} - \frac{31}{4} i

The solutions to the quadratic equation are:

u = \frac{3}{4} + i \frac{31}{4}, u = \frac{3}{4} - i \frac{31}{4}

Substitute back

u = cot (θ)

cot (θ) = \frac{3}{4} + i \frac{31}{4}, cot (θ) = \frac{3}{4} - i \frac{31}{4}

cot (θ) = \frac{3}{4} + i \frac{31}{4}, cot (θ) = \frac{3}{4} - i \frac{31}{4}

cot (θ) = \frac{3}{4} + i \frac{31}{4} :

No Solution

cot (θ) = \frac{3}{4} + i \frac{31}{4}

No Solution

cot (θ) = \frac{3}{4} - i \frac{31}{4} :

No Solution

cot (θ) = \frac{3}{4} - i \frac{31}{4}

No Solution

Combine all the solutions

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 2csc^2(θ)-3cot(θ)+3=0 ?
The general solution for 2csc^2(θ)-3cot(θ)+3=0 is No Solution for θ\in\mathbb{R}