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Popular Trigonometry >

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

  • Pre Algebra
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Solution

6tan6(x)1​=6sec6(x)1​

Solution

NoSolutionforx∈R
Solution steps
6tan6(x)1​=6sec6(x)1​
Subtract 6sec6(x)1​ from both sides6tan6(x)1​−6sec6(x)1​=0
Simplify 6tan6(x)1​−6sec6(x)1​:6tan6(x)sec6(x)sec6(x)−tan6(x)​
6tan6(x)1​−6sec6(x)1​
Least Common Multiplier of 6tan6(x),6sec6(x):6tan6(x)sec6(x)
6tan6(x),6sec6(x)
Lowest Common Multiplier (LCM)
Least Common Multiplier of 6,6:6
6,6
Least Common Multiplier (LCM)
Prime factorization of 6:2⋅3
6
6divides by 26=3⋅2=2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅3
Prime factorization of 6:2⋅3
6
6divides by 26=3⋅2=2⋅3
2,3 are all prime numbers, therefore no further factorization is possible=2⋅3
Multiply each factor the greatest number of times it occurs in either 6 or 6=2⋅3
Multiply the numbers: 2⋅3=6=6
Compute an expression comprised of factors that appear either in 6tan6(x) or 6sec6(x)=6tan6(x)sec6(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 6tan6(x)sec6(x)
For 6tan6(x)1​:multiply the denominator and numerator by sec6(x)6tan6(x)1​=6tan6(x)sec6(x)1⋅sec6(x)​=6tan6(x)sec6(x)sec6(x)​
For 6sec6(x)1​:multiply the denominator and numerator by tan6(x)6sec6(x)1​=6sec6(x)tan6(x)1⋅tan6(x)​=6tan6(x)sec6(x)tan6(x)​
=6tan6(x)sec6(x)sec6(x)​−6tan6(x)sec6(x)tan6(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=6tan6(x)sec6(x)sec6(x)−tan6(x)​
6tan6(x)sec6(x)sec6(x)−tan6(x)​=0
g(x)f(x)​=0⇒f(x)=0sec6(x)−tan6(x)=0
Factor sec6(x)−tan6(x):(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))(sec(x)−tan(x))(sec2(x)+sec(x)tan(x)+tan2(x))
sec6(x)−tan6(x)
Rewrite sec6(x)−tan6(x) as (sec3(x))2−(tan3(x))2
sec6(x)−tan6(x)
Apply exponent rule: abc=(ab)csec6(x)=(sec3(x))2=(sec3(x))2−tan6(x)
Apply exponent rule: abc=(ab)ctan6(x)=(tan3(x))2=(sec3(x))2−(tan3(x))2
=(sec3(x))2−(tan3(x))2
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)(sec3(x))2−(tan3(x))2=(sec3(x)+tan3(x))(sec3(x)−tan3(x))=(sec3(x)+tan3(x))(sec3(x)−tan3(x))
Factor sec3(x)+tan3(x):(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))
sec3(x)+tan3(x)
Apply Sum of Cubes Formula: x3+y3=(x+y)(x2−xy+y2)sec3(x)+tan3(x)=(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))=(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))
=(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))(sec3(x)−tan3(x))
Factor sec3(x)−tan3(x):(sec(x)−tan(x))(sec2(x)+sec(x)tan(x)+tan2(x))
sec3(x)−tan3(x)
Apply Difference of Cubes Formula: x3−y3=(x−y)(x2+xy+y2)sec3(x)−tan3(x)=(sec(x)−tan(x))(sec2(x)+sec(x)tan(x)+tan2(x))=(sec(x)−tan(x))(sec2(x)+sec(x)tan(x)+tan2(x))
=(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))(sec(x)−tan(x))(sec2(x)+sec(x)tan(x)+tan2(x))
(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))(sec(x)−tan(x))(sec2(x)+sec(x)tan(x)+tan2(x))=0
Rewrite using trig identities
(sec(x)+tan(x))(sec2(x)−sec(x)tan(x)+tan2(x))(sec(x)−tan(x))(sec2(x)+sec(x)tan(x)+tan2(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)):sec2(x)−tan2(x)
(sec(x)+tan(x))(sec(x)−tan(x))
Apply Difference of Two Squares Formula: (a+b)(a−b)=a2−b2a=sec(x),b=tan(x)=sec2(x)−tan2(x)
=sec2(x)−tan2(x)
Use the Pythagorean identity: sec2(x)=tan2(x)+1sec2(x)−tan2(x)=1=1
=1⋅(sec2(x)+tan2(x)+sec(x)tan(x))(sec2(x)+tan2(x)−sec(x)tan(x))
Simplify 1⋅(sec2(x)+tan2(x)+sec(x)tan(x))(sec2(x)+tan2(x)−sec(x)tan(x)):(sec2(x)+tan2(x)+sec(x)tan(x))(sec2(x)+tan2(x)−sec(x)tan(x))
1⋅(sec2(x)+tan2(x)+sec(x)tan(x))(sec2(x)+tan2(x)−sec(x)tan(x))
Multiply: 1⋅(sec2(x)+tan2(x)+sec(x)tan(x))=(sec2(x)+tan2(x)+sec(x)tan(x))=(sec2(x)+tan2(x)+sec(x)tan(x))(sec2(x)+tan2(x)−sec(x)tan(x))
=(sec2(x)+tan2(x)+sec(x)tan(x))(sec2(x)+tan2(x)−sec(x)tan(x))
(sec2(x)+tan2(x)+sec(x)tan(x))(sec2(x)+tan2(x)−sec(x)tan(x))=0
Solving each part separatelysec2(x)+tan2(x)+sec(x)tan(x)=0orsec2(x)+tan2(x)−sec(x)tan(x)=0
sec2(x)+tan2(x)+sec(x)tan(x)=0:No Solution
sec2(x)+tan2(x)+sec(x)tan(x)=0
Express with sin, cos
sec2(x)+tan2(x)+sec(x)tan(x)
Use the basic trigonometric identity: sec(x)=cos(x)1​=(cos(x)1​)2+tan2(x)+cos(x)1​tan(x)
Use the basic trigonometric identity: tan(x)=cos(x)sin(x)​=(cos(x)1​)2+(cos(x)sin(x)​)2+cos(x)1​⋅cos(x)sin(x)​
Simplify (cos(x)1​)2+(cos(x)sin(x)​)2+cos(x)1​⋅cos(x)sin(x)​:cos2(x)1+sin2(x)+sin(x)​
(cos(x)1​)2+(cos(x)sin(x)​)2+cos(x)1​⋅cos(x)sin(x)​
(cos(x)1​)2=cos2(x)1​
(cos(x)1​)2
Apply exponent rule: (ba​)c=bcac​=cos2(x)12​
Apply rule 1a=112=1=cos2(x)1​
(cos(x)sin(x)​)2=cos2(x)sin2(x)​
(cos(x)sin(x)​)2
Apply exponent rule: (ba​)c=bcac​=cos2(x)sin2(x)​
cos(x)1​⋅cos(x)sin(x)​=cos2(x)sin(x)​
cos(x)1​⋅cos(x)sin(x)​
Multiply fractions: ba​⋅dc​=b⋅da⋅c​=cos(x)cos(x)1⋅sin(x)​
Multiply: 1⋅sin(x)=sin(x)=cos(x)cos(x)sin(x)​
cos(x)cos(x)=cos2(x)
cos(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=cos1+1(x)
Add the numbers: 1+1=2=cos2(x)
=cos2(x)sin(x)​
=cos2(x)1​+cos2(x)sin2(x)​+cos2(x)sin(x)​
Apply rule ca​±cb​=ca±b​=cos2(x)1+sin2(x)+sin(x)​
=cos2(x)1+sin2(x)+sin(x)​
cos2(x)1+sin(x)+sin2(x)​=0
g(x)f(x)​=0⇒f(x)=01+sin(x)+sin2(x)=0
Solve by substitution
1+sin(x)+sin2(x)=0
Let: sin(x)=u1+u+u2=0
1+u+u2=0:u=−21​+i23​​,u=−21​−i23​​
1+u+u2=0
Write in the standard form ax2+bx+c=0u2+u+1=0
Solve with the quadratic formula
u2+u+1=0
Quadratic Equation Formula:
For a=1,b=1,c=1u1,2​=2⋅1−1±12−4⋅1⋅1​​
u1,2​=2⋅1−1±12−4⋅1⋅1​​
Simplify 12−4⋅1⋅1​:3​i
12−4⋅1⋅1​
Apply rule 1a=112=1=1−4⋅1⋅1​
Multiply the numbers: 4⋅1⋅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
u1,2​=2⋅1−1±3​i​
Separate the solutionsu1​=2⋅1−1+3​i​,u2​=2⋅1−1−3​i​
u=2⋅1−1+3​i​:−21​+i23​​
2⋅1−1+3​i​
Multiply the numbers: 2⋅1=2=2−1+3​i​
Rewrite 2−1+3​i​ in standard complex form: −21​+23​​i
2−1+3​i​
Apply the fraction rule: ca±b​=ca​±cb​2−1+3​i​=−21​+23​i​=−21​+23​i​
=−21​+23​​i
u=2⋅1−1−3​i​:−21​−i23​​
2⋅1−1−3​i​
Multiply the numbers: 2⋅1=2=2−1−3​i​
Rewrite 2−1−3​i​ in standard complex form: −21​−23​​i
2−1−3​i​
Apply the fraction rule: ca±b​=ca​±cb​2−1−3​i​=−21​−23​i​=−21​−23​i​
=−21​−23​​i
The solutions to the quadratic equation are:u=−21​+i23​​,u=−21​−i23​​
Substitute back u=sin(x)sin(x)=−21​+i23​​,sin(x)=−21​−i23​​
sin(x)=−21​+i23​​,sin(x)=−21​−i23​​
sin(x)=−21​+i23​​:No Solution
sin(x)=−21​+i23​​
NoSolution
sin(x)=−21​−i23​​:No Solution
sin(x)=−21​−i23​​
NoSolution
Combine all the solutionsNoSolution
sec2(x)+tan2(x)−sec(x)tan(x)=0:No Solution
sec2(x)+tan2(x)−sec(x)tan(x)=0
Express with sin, cos
sec2(x)+tan2(x)−sec(x)tan(x)
Use the basic trigonometric identity: sec(x)=cos(x)1​=(cos(x)1​)2+tan2(x)−cos(x)1​tan(x)
Use the basic trigonometric identity: tan(x)=cos(x)sin(x)​=(cos(x)1​)2+(cos(x)sin(x)​)2−cos(x)1​⋅cos(x)sin(x)​
Simplify (cos(x)1​)2+(cos(x)sin(x)​)2−cos(x)1​⋅cos(x)sin(x)​:cos2(x)1+sin2(x)−sin(x)​
(cos(x)1​)2+(cos(x)sin(x)​)2−cos(x)1​⋅cos(x)sin(x)​
(cos(x)1​)2=cos2(x)1​
(cos(x)1​)2
Apply exponent rule: (ba​)c=bcac​=cos2(x)12​
Apply rule 1a=112=1=cos2(x)1​
(cos(x)sin(x)​)2=cos2(x)sin2(x)​
(cos(x)sin(x)​)2
Apply exponent rule: (ba​)c=bcac​=cos2(x)sin2(x)​
cos(x)1​⋅cos(x)sin(x)​=cos2(x)sin(x)​
cos(x)1​⋅cos(x)sin(x)​
Multiply fractions: ba​⋅dc​=b⋅da⋅c​=cos(x)cos(x)1⋅sin(x)​
Multiply: 1⋅sin(x)=sin(x)=cos(x)cos(x)sin(x)​
cos(x)cos(x)=cos2(x)
cos(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=cos1+1(x)
Add the numbers: 1+1=2=cos2(x)
=cos2(x)sin(x)​
=cos2(x)1​+cos2(x)sin2(x)​−cos2(x)sin(x)​
Apply rule ca​±cb​=ca±b​=cos2(x)1+sin2(x)−sin(x)​
=cos2(x)1+sin2(x)−sin(x)​
cos2(x)1−sin(x)+sin2(x)​=0
g(x)f(x)​=0⇒f(x)=01−sin(x)+sin2(x)=0
Solve by substitution
1−sin(x)+sin2(x)=0
Let: sin(x)=u1−u+u2=0
1−u+u2=0:u=21​+i23​​,u=21​−i23​​
1−u+u2=0
Write in the standard form ax2+bx+c=0u2−u+1=0
Solve with the quadratic formula
u2−u+1=0
Quadratic Equation Formula:
For a=1,b=−1,c=1u1,2​=2⋅1−(−1)±(−1)2−4⋅1⋅1​​
u1,2​=2⋅1−(−1)±(−1)2−4⋅1⋅1​​
Simplify (−1)2−4⋅1⋅1​:3​i
(−1)2−4⋅1⋅1​
(−1)2=1
(−1)2
Apply exponent rule: (−a)n=an,if n is even(−1)2=12=12
Apply rule 1a=1=1
4⋅1⋅1=4
4⋅1⋅1
Multiply the numbers: 4⋅1⋅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
u1,2​=2⋅1−(−1)±3​i​
Separate the solutionsu1​=2⋅1−(−1)+3​i​,u2​=2⋅1−(−1)−3​i​
u=2⋅1−(−1)+3​i​:21​+i23​​
2⋅1−(−1)+3​i​
Apply rule −(−a)=a=2⋅11+3​i​
Multiply the numbers: 2⋅1=2=21+3​i​
Rewrite 21+3​i​ in standard complex form: 21​+23​​i
21+3​i​
Apply the fraction rule: ca±b​=ca​±cb​21+3​i​=21​+23​i​=21​+23​i​
=21​+23​​i
u=2⋅1−(−1)−3​i​:21​−i23​​
2⋅1−(−1)−3​i​
Apply rule −(−a)=a=2⋅11−3​i​
Multiply the numbers: 2⋅1=2=21−3​i​
Rewrite 21−3​i​ in standard complex form: 21​−23​​i
21−3​i​
Apply the fraction rule: ca±b​=ca​±cb​21−3​i​=21​−23​i​=21​−23​i​
=21​−23​​i
The solutions to the quadratic equation are:u=21​+i23​​,u=21​−i23​​
Substitute back u=sin(x)sin(x)=21​+i23​​,sin(x)=21​−i23​​
sin(x)=21​+i23​​,sin(x)=21​−i23​​
sin(x)=21​+i23​​:No Solution
sin(x)=21​+i23​​
NoSolution
sin(x)=21​−i23​​:No Solution
sin(x)=21​−i23​​
NoSolution
Combine all the solutionsNoSolution
Combine all the solutionsNoSolutionforx∈R

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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}
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