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

The general solution for 3csc^2(5x)=-4 is No Solution for x\in\mathbb{R}

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3csc^2(5x)=-4

Solution

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

Solution

No Solution for x \in R

Solution steps

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

Solve by substitution

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

Let:

csc (5 x) = u

3 u^{2} = - 4

3 u^{2} = - 4 : u = i \frac{2 3}{3}, u = - i \frac{2 3}{3}

3 u^{2} = - 4

Divide both sides by

3

3 u^{2} = - 4

Divide both sides by

3

\frac{3 u ^{2}}{3} = \frac{- 4}{3}

Simplify

u^{2} = - \frac{4}{3}

u^{2} = - \frac{4}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - \frac{4}{3}, u = - - \frac{4}{3}

Simplify

- \frac{4}{3} : i \frac{2 3}{3}

- \frac{4}{3}

Apply radical rule:

- a = - 1 a

- \frac{4}{3} = - 1 \frac{4}{3}

= - 1 \frac{4}{3}

Apply imaginary number rule:

- 1 = i

= i \frac{4}{3}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{4}{3} = \frac{4}{3}

= i \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{2}{3}

\frac{2}{3} = \frac{2 3}{3}

\frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

= i \frac{2 3}{3}

Rewrite

i \frac{2 3}{3}

in standard complex form:

\frac{2 3}{3} i

i \frac{2 3}{3}

\frac{2 3}{3} = \frac{2}{3}

\frac{2 3}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{2 \cdot 3 ^{\frac{1}{2}}}{3}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{2}{3 ^{1 - \frac{1}{2}}}

Subtract the numbers:

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

= \frac{2}{3 ^{\frac{1}{2}}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= \frac{2}{3}

= i \frac{2}{3}

Multiply fractions:

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

= \frac{2 i}{3}

\frac{2}{3} = \frac{2 3}{3}

\frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

= \frac{2 3}{3} i

= \frac{2 3}{3} i

Simplify

- - \frac{4}{3} : - i \frac{2 3}{3}

- - \frac{4}{3}

Simplify

- \frac{4}{3} : i \frac{2}{3}

- \frac{4}{3}

Apply radical rule:

- a = - 1 a

- \frac{4}{3} = - 1 \frac{4}{3}

= - 1 \frac{4}{3}

Apply imaginary number rule:

- 1 = i

= i \frac{4}{3}

Apply radical rule:

assuming

a \geq 0, b \geq 0

\frac{4}{3} = \frac{4}{3}

= i \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= i \frac{2}{3}

= - i \frac{2}{3}

\frac{2}{3} = \frac{2 3}{3}

\frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

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

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

Substitute back

u = csc (5 x)

csc (5 x) = i \frac{2 3}{3}, csc (5 x) = - i \frac{2 3}{3}

csc (5 x) = i \frac{2 3}{3}, csc (5 x) = - i \frac{2 3}{3}

csc (5 x) = i \frac{2 3}{3} :

No Solution

csc (5 x) = i \frac{2 3}{3}

No Solution

csc (5 x) = - i \frac{2 3}{3} :

No Solution

csc (5 x) = - i \frac{2 3}{3}

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 3csc^2(5x)=-4 ?
The general solution for 3csc^2(5x)=-4 is No Solution for x\in\mathbb{R}