What is the general solution for 2csc^2(x)=cot(x) ?

The general solution for 2csc^2(x)=cot(x) is No Solution for x\in\mathbb{R}

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Popular >

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

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Subtract

cot (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

- cot (x) + (1 + cot^{2} (x)) \cdot 2 = 0

Solve by substitution

- cot (x) + (1 + cot^{2} (x)) \cdot 2 = 0

Let:

cot (x) = u

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

- u + (1 + u^{2}) \cdot 2 = 0 : u = \frac{1}{4} + i \frac{15}{4}, u = \frac{1}{4} - i \frac{15}{4}

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

Expand

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

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

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

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 \cdot 1 + 2 u^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 + 2 u^{2}

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 1, c = 2

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

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 2 : 15 i

(- 1)^{2} - 4 \cdot 2 \cdot 2

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 - 16

Subtract the numbers:

1 - 16 = - 15

= - 15

Apply radical rule:

- a = - 1 a

- 15 = - 1 15

= - 1 15

Apply imaginary number rule:

- 1 = i

= 15 i

u_{1, 2} = \frac{- ( - 1 ) \pm 15 i}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 15 i}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 15 i}{2 \cdot 2}

u = \frac{- ( - 1 ) + 15 i}{2 \cdot 2} : \frac{1}{4} + i \frac{15}{4}

\frac{- ( - 1 ) + 15 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 15 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 15 i}{4}

Rewrite

\frac{1 + 15 i}{4}

in standard complex form:

\frac{1}{4} + \frac{15}{4} i

\frac{1 + 15 i}{4}

Apply the fraction rule:

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

\frac{1 + 15 i}{4} = \frac{1}{4} + \frac{15 i}{4}

= \frac{1}{4} + \frac{15 i}{4}

= \frac{1}{4} + \frac{15}{4} i

u = \frac{- ( - 1 ) - 15 i}{2 \cdot 2} : \frac{1}{4} - i \frac{15}{4}

\frac{- ( - 1 ) - 15 i}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 15 i}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 15 i}{4}

Rewrite

\frac{1 - 15 i}{4}

in standard complex form:

\frac{1}{4} - \frac{15}{4} i

\frac{1 - 15 i}{4}

Apply the fraction rule:

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

\frac{1 - 15 i}{4} = \frac{1}{4} - \frac{15 i}{4}

= \frac{1}{4} - \frac{15 i}{4}

= \frac{1}{4} - \frac{15}{4} i

The solutions to the quadratic equation are:

u = \frac{1}{4} + i \frac{15}{4}, u = \frac{1}{4} - i \frac{15}{4}

Substitute back

u = cot (x)

cot (x) = \frac{1}{4} + i \frac{15}{4}, cot (x) = \frac{1}{4} - i \frac{15}{4}

cot (x) = \frac{1}{4} + i \frac{15}{4}, cot (x) = \frac{1}{4} - i \frac{15}{4}

cot (x) = \frac{1}{4} + i \frac{15}{4} :

No Solution

cot (x) = \frac{1}{4} + i \frac{15}{4}

No Solution

cot (x) = \frac{1}{4} - i \frac{15}{4} :

No Solution

cot (x) = \frac{1}{4} - i \frac{15}{4}

No Solution

Combine all the solutions

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 2csc^2(x)=cot(x) ?
The general solution for 2csc^2(x)=cot(x) is No Solution for x\in\mathbb{R}