What is the general solution for (tan^2(3x))(csc(3x))+1=-1*sec^2(3x) ?

The general solution for (tan^2(3x))(csc(3x))+1=-1*sec^2(3x) is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(tan^2(3x))(csc(3x))+1=-1*sec^2(3x)

Solution

(tan^{2} (3 x)) (csc (3 x)) + 1 = - 1 \cdot sec^{2} (3 x)

Solution

No Solution for x \in R

Solution steps

(tan^{2} (3 x)) (csc (3 x)) + 1 = - 1 \cdot sec^{2} (3 x)

Subtract

- 1 sec^{2} (3 x)

from both sides

tan^{2} (3 x) csc (3 x) + 1 + sec^{2} (3 x) = 0

Express with sin, cos

(\frac{sin ( 3 x )}{cos ( 3 x )})^{2} \frac{1}{sin ( 3 x )} + 1 + (\frac{1}{cos ( 3 x )})^{2} = 0

Simplify

(\frac{sin ( 3 x )}{cos ( 3 x )})^{2} \frac{1}{sin ( 3 x )} + 1 + (\frac{1}{cos ( 3 x )})^{2} : \frac{sin ( 3 x ) + 1 + cos ^{2} ( 3 x )}{cos ^{2} ( 3 x )}

(\frac{sin ( 3 x )}{cos ( 3 x )})^{2} \frac{1}{sin ( 3 x )} + 1 + (\frac{1}{cos ( 3 x )})^{2}

(\frac{sin ( 3 x )}{cos ( 3 x )})^{2} \frac{1}{sin ( 3 x )} = \frac{sin ( 3 x )}{cos ^{2} ( 3 x )}

(\frac{sin ( 3 x )}{cos ( 3 x )})^{2} \frac{1}{sin ( 3 x )}

Multiply fractions:

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

= \frac{1 \cdot ( \frac{s i n ( 3 x )}{c o s ( 3 x )} ) ^{2}}{sin ( 3 x )}

Multiply:

1 \cdot (\frac{sin ( 3 x )}{cos ( 3 x )})^{2} = (\frac{sin ( 3 x )}{cos ( 3 x )})^{2}

= \frac{( \frac{s i n ( 3 x )}{c o s ( 3 x )} ) ^{2}}{sin ( 3 x )}

(\frac{sin ( 3 x )}{cos ( 3 x )})^{2} = \frac{sin ^{2} ( 3 x )}{cos ^{2} ( 3 x )}

(\frac{sin ( 3 x )}{cos ( 3 x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{sin ^{2} ( 3 x )}{cos ^{2} ( 3 x )}

= \frac{\frac{s i n ^{2} ( 3 x )}{c o s ^{2} ( 3 x )}}{sin ( 3 x )}

Apply the fraction rule:

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

= \frac{sin ^{2} ( 3 x )}{cos ^{2} ( 3 x ) sin ( 3 x )}

Cancel the common factor:

sin (3 x)

= \frac{sin ( 3 x )}{cos ^{2} ( 3 x )}

(\frac{1}{cos ( 3 x )})^{2} = \frac{1}{cos ^{2} ( 3 x )}

(\frac{1}{cos ( 3 x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( 3 x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( 3 x )}

= \frac{sin ( 3 x )}{cos ^{2} ( 3 x )} + 1 + \frac{1}{cos ^{2} ( 3 x )}

Combine the fractions

\frac{sin ( 3 x )}{cos ^{2} ( 3 x )} + \frac{1}{cos ^{2} ( 3 x )} : \frac{sin ( 3 x ) + 1}{cos ^{2} ( 3 x )}

Apply rule

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

= \frac{sin ( 3 x ) + 1}{cos ^{2} ( 3 x )}

= \frac{sin ( 3 x ) + 1}{cos ^{2} ( 3 x )} + 1

Convert element to fraction:

1 = \frac{1 cos ^{2} ( 3 x )}{cos ^{2} ( 3 x )}

= \frac{sin ( 3 x ) + 1}{cos ^{2} ( 3 x )} + \frac{1 \cdot cos ^{2} ( 3 x )}{cos ^{2} ( 3 x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( 3 x ) + 1 + 1 \cdot cos ^{2} ( 3 x )}{cos ^{2} ( 3 x )}

Multiply:

1 \cdot cos^{2} (3 x) = cos^{2} (3 x)

= \frac{sin ( 3 x ) + 1 + cos ^{2} ( 3 x )}{cos ^{2} ( 3 x )}

\frac{sin ( 3 x ) + 1 + cos ^{2} ( 3 x )}{cos ^{2} ( 3 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (3 x) + 1 + cos^{2} (3 x) = 0

Subtract

cos^{2} (3 x)

from both sides

sin (3 x) + 1 = - cos^{2} (3 x)

Square both sides

(sin (3 x) + 1)^{2} = (- cos^{2} (3 x))^{2}

Subtract

(- cos^{2} (3 x))^{2}

from both sides

(sin (3 x) + 1)^{2} - cos^{4} (3 x) = 0

Factor

(sin (3 x) + 1)^{2} - cos^{4} (3 x) : (sin (3 x) + 1 + cos^{2} (3 x)) (sin (3 x) + 1 - cos^{2} (3 x))

(sin (3 x) + 1)^{2} - cos^{4} (3 x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

cos^{4} (3 x) = (cos^{2} (3 x))^{2}

= (sin (3 x) + 1)^{2} - (cos^{2} (3 x))^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(sin (3 x) + 1)^{2} - (cos^{2} (3 x))^{2} = ((sin (3 x) + 1) + cos^{2} (3 x)) ((sin (3 x) + 1) - cos^{2} (3 x))

= ((sin (3 x) + 1) + cos^{2} (3 x)) ((sin (3 x) + 1) - cos^{2} (3 x))

Refine

= (cos^{2} (3 x) + sin (3 x) + 1) (sin (3 x) - cos^{2} (3 x) + 1)

(sin (3 x) + 1 + cos^{2} (3 x)) (sin (3 x) + 1 - cos^{2} (3 x)) = 0

Solving each part separately

sin (3 x) + 1 + cos^{2} (3 x) = 0 or sin (3 x) + 1 - cos^{2} (3 x) = 0

sin (3 x) + 1 + cos^{2} (3 x) = 0 : x = \frac{π}{2} + \frac{2 πn}{3}

sin (3 x) + 1 + cos^{2} (3 x) = 0

Rewrite using trig identities

1 + cos^{2} (3 x) + sin (3 x)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 1 + 1 - sin^{2} (3 x) + sin (3 x)

Simplify

= sin (3 x) - sin^{2} (3 x) + 2

2 + sin (3 x) - sin^{2} (3 x) = 0

Solve by substitution

2 + sin (3 x) - sin^{2} (3 x) = 0

Let:

sin (3 x) = u

2 + u - u^{2} = 0

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

2 + u - u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

1^{2} - 4 (- 1) \cdot 2 = 3

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 2

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

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

Separate the solutions

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

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

\frac{- 1 + 3}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 1}

Add/Subtract the numbers:

- 1 + 3 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

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

\frac{- 1 - 3}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 3 = - 4

= \frac{- 4}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{- 2}

Apply the fraction rule:

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

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

The solutions to the quadratic equation are:

u = - 1, u = 2

Substitute back

u = sin (3 x)

sin (3 x) = - 1, sin (3 x) = 2

sin (3 x) = - 1, sin (3 x) = 2

sin (3 x) = - 1 : x = \frac{π}{2} + \frac{2 πn}{3}

sin (3 x) = - 1

General solutions for

sin (3 x) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

3 x = \frac{3 π}{2} + 2 πn

3 x = \frac{3 π}{2} + 2 πn

Solve

3 x = \frac{3 π}{2} + 2 πn : x = \frac{π}{2} + \frac{2 πn}{3}

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{3 π}{2}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{3 π}{2}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

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

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

sin (3 x) = 2 :

No Solution

sin (3 x) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + \frac{2 πn}{3}

sin (3 x) + 1 - cos^{2} (3 x) = 0 : x = \frac{2 πn}{3}, x = \frac{π}{3} + \frac{2 πn}{3}, x = \frac{π}{2} + \frac{2 πn}{3}

sin (3 x) + 1 - cos^{2} (3 x) = 0

Rewrite using trig identities

1 - cos^{2} (3 x) + sin (3 x)

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - cos^{2} (x) = sin^{2} (x)

= sin (3 x) + sin^{2} (3 x)

sin (3 x) + sin^{2} (3 x) = 0

Solve by substitution

sin (3 x) + sin^{2} (3 x) = 0

Let:

sin (3 x) = u

u + u^{2} = 0

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

u + u^{2} = 0

Write in the standard form

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

u^{2} + u = 0

Solve with the quadratic formula

u^{2} + u = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot 0 = 1

1^{2} - 4 \cdot 1 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- 1 + 1}{2 \cdot 1} : 0

\frac{- 1 + 1}{2 \cdot 1}

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 1 - 1}{2 \cdot 1} : - 1

\frac{- 1 - 1}{2 \cdot 1}

Subtract the numbers:

- 1 - 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 0, u = - 1

Substitute back

u = sin (3 x)

sin (3 x) = 0, sin (3 x) = - 1

sin (3 x) = 0, sin (3 x) = - 1

sin (3 x) = 0 : x = \frac{2 πn}{3}, x = \frac{π}{3} + \frac{2 πn}{3}

sin (3 x) = 0

General solutions for

sin (3 x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

3 x = 0 + 2 πn, 3 x = π + 2 πn

3 x = 0 + 2 πn, 3 x = π + 2 πn

Solve

3 x = 0 + 2 πn : x = \frac{2 πn}{3}

3 x = 0 + 2 πn

0 + 2 πn = 2 πn

3 x = 2 πn

Divide both sides by

3

3 x = 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{2 πn}{3}

Simplify

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

Solve

3 x = π + 2 πn : x = \frac{π}{3} + \frac{2 πn}{3}

3 x = π + 2 πn

Divide both sides by

3

3 x = π + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{π}{3} + \frac{2 πn}{3}

Simplify

x = \frac{π}{3} + \frac{2 πn}{3}

x = \frac{π}{3} + \frac{2 πn}{3}

x = \frac{2 πn}{3}, x = \frac{π}{3} + \frac{2 πn}{3}

sin (3 x) = - 1 : x = \frac{π}{2} + \frac{2 πn}{3}

sin (3 x) = - 1

General solutions for

sin (3 x) = - 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

3 x = \frac{3 π}{2} + 2 πn

3 x = \frac{3 π}{2} + 2 πn

Solve

3 x = \frac{3 π}{2} + 2 πn : x = \frac{π}{2} + \frac{2 πn}{3}

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

3 x = \frac{3 π}{2} + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{3 π}{2}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{3 π}{2}}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

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

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

\frac{\frac{3 π}{2}}{3} = \frac{π}{2}

\frac{\frac{3 π}{2}}{3}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3 π}{6}

Cancel the common factor:

3

= \frac{π}{2}

= \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

x = \frac{π}{2} + \frac{2 πn}{3}

Combine all the solutions

x = \frac{2 πn}{3}, x = \frac{π}{3} + \frac{2 πn}{3}, x = \frac{π}{2} + \frac{2 πn}{3}

Combine all the solutions

x = \frac{π}{2} + \frac{2 πn}{3}, x = \frac{2 πn}{3}, x = \frac{π}{3} + \frac{2 πn}{3}

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

tan^{2} (3 x) csc (3 x) + 1 = - 1 sec^{2} (3 x)

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + \frac{2 πn}{3} :

True

\frac{π}{2} + \frac{2 πn}{3}

Plug in

n = 1

\frac{π}{2} + \frac{2 π 1}{3}

For

tan^{2} (3 x) csc (3 x) + 1 = - 1 sec^{2} (3 x)

plug in

x = \frac{π}{2} + \frac{2 π 1}{3}

tan^{2} (3 (\frac{π}{2} + \frac{2 π 1}{3})) csc (3 (\frac{π}{2} + \frac{2 π 1}{3})) + 1 = - 1 \cdot sec^{2} (3 (\frac{π}{2} + \frac{2 π 1}{3}))

Refine

- \infty = - \infty

\Rightarrow True

Check the solution

\frac{2 πn}{3} :

False

\frac{2 πn}{3}

Plug in

n = 1

\frac{2 π 1}{3}

For

tan^{2} (3 x) csc (3 x) + 1 = - 1 sec^{2} (3 x)

plug in

x = \frac{2 π 1}{3}

tan^{2} (3 \cdot \frac{2 π 1}{3}) csc (3 \cdot \frac{2 π 1}{3}) + 1 = - 1 \cdot sec^{2} (3 \cdot \frac{2 π 1}{3})

Undefined

\Rightarrow False

Check the solution

\frac{π}{3} + \frac{2 πn}{3} :

False

\frac{π}{3} + \frac{2 πn}{3}

Plug in

n = 1

\frac{π}{3} + \frac{2 π 1}{3}

For

tan^{2} (3 x) csc (3 x) + 1 = - 1 sec^{2} (3 x)

plug in

x = \frac{π}{3} + \frac{2 π 1}{3}

tan^{2} (3 (\frac{π}{3} + \frac{2 π 1}{3})) csc (3 (\frac{π}{3} + \frac{2 π 1}{3})) + 1 = - 1 \cdot sec^{2} (3 (\frac{π}{3} + \frac{2 π 1}{3}))

Undefined

\Rightarrow False

x = \frac{π}{2} + \frac{2 πn}{3}

Since the equation is undefined for:

\frac{π}{2} + \frac{2 πn}{3}

No Solution for x \in R

Popular Examples

(4sin(x)-3cos(x))/(3sin(x)+4cos(x))=3

\frac{4 sin ( x ) - 3 cos ( x )}{3 sin ( x ) + 4 cos ( x )} = 3

7=10sin(40(0-h))+17

7 = 10 sin (40 (0 - h)) + 17

sin(2x)=cos(x),0<= x<2pi

sin (2 x) = cos (x), 0 \leq x < 2 π

sec(x)=-5/3

sec (x) = - \frac{5}{3}

(sin(x))/(1+cos(x))=sin(x/2)

\frac{sin ( x )}{1 + cos ( x )} = sin (\frac{x}{2})

Frequently Asked Questions (FAQ)

What is the general solution for (tan^2(3x))(csc(3x))+1=-1*sec^2(3x) ?
The general solution for (tan^2(3x))(csc(3x))+1=-1*sec^2(3x) is No Solution for x\in\mathbb{R}