What is the general solution for 1/(6tan^6(x))= 1/(6sec^6(x)) ?

The general solution for 1/(6tan^6(x))= 1/(6sec^6(x)) is No Solution for x\in\mathbb{R}

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1/(6tan^6(x))= 1/(6sec^6(x))

Solution

\frac{1}{6 tan ^{6} ( x )} = \frac{1}{6 sec ^{6} ( x )}

Solution

No Solution for x \in R

Solution steps

\frac{1}{6 tan ^{6} ( x )} = \frac{1}{6 sec ^{6} ( x )}

Subtract

\frac{1}{6 sec ^{6} ( x )}

from both sides

\frac{1}{6 tan ^{6} ( x )} - \frac{1}{6 sec ^{6} ( x )} = 0

Simplify

\frac{1}{6 tan ^{6} ( x )} - \frac{1}{6 sec ^{6} ( x )} : \frac{sec ^{6} ( x ) - tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

\frac{1}{6 tan ^{6} ( x )} - \frac{1}{6 sec ^{6} ( x )}

Least Common Multiplier of

6 tan^{6} (x), 6 sec^{6} (x) : 6 tan^{6} (x) sec^{6} (x)

6 tan^{6} (x), 6 sec^{6} (x)

Lowest Common Multiplier (LCM)

Least Common Multiplier of

6, 6 : 6

6, 6

Least Common Multiplier (LCM)

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

6

6

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

Compute an expression comprised of factors that appear either in

6 tan^{6} (x)

6 sec^{6} (x)

= 6 tan^{6} (x) sec^{6} (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6 tan^{6} (x) sec^{6} (x)

For

\frac{1}{6 tan ^{6} ( x )} :

multiply the denominator and numerator by

sec^{6} (x)

\frac{1}{6 tan ^{6} ( x )} = \frac{1 \cdot sec ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )} = \frac{sec ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

For

\frac{1}{6 sec ^{6} ( x )} :

multiply the denominator and numerator by

tan^{6} (x)

\frac{1}{6 sec ^{6} ( x )} = \frac{1 \cdot tan ^{6} ( x )}{6 sec ^{6} ( x ) tan ^{6} ( x )} = \frac{tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

= \frac{sec ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )} - \frac{tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sec ^{6} ( x ) - tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

\frac{sec ^{6} ( x ) - tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )} = 0

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

sec^{6} (x) - tan^{6} (x) = 0

Factor

sec^{6} (x) - tan^{6} (x) : (sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x)) (sec (x) - tan (x)) (sec^{2} (x) + sec (x) tan (x) + tan^{2} (x))

sec^{6} (x) - tan^{6} (x)

Rewrite

sec^{6} (x) - tan^{6} (x)

(sec^{3} (x))^{2} - (tan^{3} (x))^{2}

sec^{6} (x) - tan^{6} (x)

Apply exponent rule:

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

sec^{6} (x) = (sec^{3} (x))^{2}

= (sec^{3} (x))^{2} - tan^{6} (x)

Apply exponent rule:

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

tan^{6} (x) = (tan^{3} (x))^{2}

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

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

Apply Difference of Two Squares Formula:

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

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

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

Factor

sec^{3} (x) + tan^{3} (x) : (sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x))

sec^{3} (x) + tan^{3} (x)

Apply Sum of Cubes Formula:

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

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

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

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

Factor

sec^{3} (x) - tan^{3} (x) : (sec (x) - tan (x)) (sec^{2} (x) + sec (x) tan (x) + tan^{2} (x))

sec^{3} (x) - tan^{3} (x)

Apply Difference of Cubes Formula:

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

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

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

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

(sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x)) (sec (x) - tan (x)) (sec^{2} (x) + sec (x) tan (x) + tan^{2} (x)) = 0

Rewrite using trig identities

(sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x)) (sec (x) - tan (x)) (sec^{2} (x) + sec (x) tan (x) + tan^{2} (x))

(sec (x) + tan (x)) (sec (x) - tan (x)) = 1

(sec (x) + tan (x)) (sec (x) - tan (x))

Expand

(sec (x) + tan (x)) (sec (x) - tan (x)) : sec^{2} (x) - tan^{2} (x)

(sec (x) + tan (x)) (sec (x) - tan (x))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = sec (x), b = tan (x)

= sec^{2} (x) - tan^{2} (x)

= sec^{2} (x) - tan^{2} (x)

Use the Pythagorean identity:

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

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

= 1

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

Simplify

1 \cdot (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) (sec^{2} (x) + tan^{2} (x) - sec (x) tan (x)) : (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) (sec^{2} (x) + tan^{2} (x) - sec (x) tan (x))

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

Multiply:

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

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

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

(sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) (sec^{2} (x) + tan^{2} (x) - sec (x) tan (x)) = 0

Solving each part separately

sec^{2} (x) + tan^{2} (x) + sec (x) tan (x) = 0 or sec^{2} (x) + tan^{2} (x) - sec (x) tan (x) = 0

sec^{2} (x) + tan^{2} (x) + sec (x) tan (x) = 0 :

No Solution

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

Express with sin, cos

sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Apply exponent rule:

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

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

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

\frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

= \frac{sin ( x )}{cos ( x ) cos ( x )}

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Apply rule

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

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

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 + u + u^{2} = 0

1 + u + u^{2} = 0 : u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

1 + u + u^{2} = 0

Write in the standard form

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

u^{2} + u + 1 = 0

Solve with the quadratic formula

u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 1

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

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

Simplify

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

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

Separate the solutions

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

u = \frac{- 1 + 3 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{3}{2}

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

Multiply the numbers:

2 \cdot 1 = 2

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

Rewrite

\frac{- 1 + 3 i}{2}

in standard complex form:

- \frac{1}{2} + \frac{3}{2} i

\frac{- 1 + 3 i}{2}

Apply the fraction rule:

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

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

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

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

u = \frac{- 1 - 3 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{3}{2}

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

Multiply the numbers:

2 \cdot 1 = 2

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

Rewrite

\frac{- 1 - 3 i}{2}

in standard complex form:

- \frac{1}{2} - \frac{3}{2} i

\frac{- 1 - 3 i}{2}

Apply the fraction rule:

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

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

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

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

The solutions to the quadratic equation are:

u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

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

No Solution

sin (x) = - \frac{1}{2} + i \frac{3}{2}

No Solution

sin (x) = - \frac{1}{2} - i \frac{3}{2} :

No Solution

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

No Solution

Combine all the solutions

No Solution

sec^{2} (x) + tan^{2} (x) - sec (x) tan (x) = 0 :

No Solution

sec^{2} (x) + tan^{2} (x) - sec (x) tan (x) = 0

Express with sin, cos

sec^{2} (x) + tan^{2} (x) - sec (x) tan (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Apply exponent rule:

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

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

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

\frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

= \frac{sin ( x )}{cos ( x ) cos ( x )}

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

Apply rule

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

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

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

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 - u + u^{2} = 0

1 - u + u^{2} = 0 : u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{2}

1 - u + u^{2} = 0

Write in the standard form

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

u^{2} - u + 1 = 0

Solve with the quadratic formula

u^{2} - u + 1 = 0

Quadratic Equation Formula:

For

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

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

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

Simplify

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

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

(- 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 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 3 i}{2}

Rewrite

\frac{1 + 3 i}{2}

in standard complex form:

\frac{1}{2} + \frac{3}{2} i

\frac{1 + 3 i}{2}

Apply the fraction rule:

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

\frac{1 + 3 i}{2} = \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Rewrite

\frac{1 - 3 i}{2}

in standard complex form:

\frac{1}{2} - \frac{3}{2} i

\frac{1 - 3 i}{2}

Apply the fraction rule:

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

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

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

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

The solutions to the quadratic equation are:

u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{2}

Substitute back

u = sin (x)

sin (x) = \frac{1}{2} + i \frac{3}{2}, sin (x) = \frac{1}{2} - i \frac{3}{2}

sin (x) = \frac{1}{2} + i \frac{3}{2}, sin (x) = \frac{1}{2} - i \frac{3}{2}

sin (x) = \frac{1}{2} + i \frac{3}{2} :

No Solution

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

No Solution

sin (x) = \frac{1}{2} - i \frac{3}{2} :

No Solution

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

No Solution

Combine all the solutions

No Solution

Combine all the solutions

No Solution for x \in R

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

sin (x) cos (x) = sin (x), 0 < x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 1/(6tan^6(x))= 1/(6sec^6(x)) ?
The general solution for 1/(6tan^6(x))= 1/(6sec^6(x)) is No Solution for x\in\mathbb{R}