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Popular Trigonometry >

1/(tan(x))>cot(1/x)

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Solution

tan(x)1​>cot(x1​)

Solution

2πn<x<−29π−81π2+4​​+2πnor9π1​+2πn<x<−4π+16π2+1​+2πnor8π1​+2πn<x<−27π−49π2+4​​+2πnor7π1​+2πn<x<−3π+9π2+1​+2πnor6π1​+2πn<x<−25π−25π2+4​​+2πnor5π1​+2πn<x<−2π+4π2+1​+2πnor4π1​+2πn<x<−23π−9π2+4​​+2πnor3π1​+2πn<x<−π+π2+1​+2πnor2π1​+2πn<x<−2π−π2+4​​+2πnorπ1​+2πn<x<1+2πnorπ+2πn<x<−2−π−π2+4​​+2πn
+2
Interval Notation
(2πn,−29π−81π2+4​​+2πn)∪(9π1​+2πn,−4π+16π2+1​+2πn)∪(8π1​+2πn,−27π−49π2+4​​+2πn)∪(7π1​+2πn,−3π+9π2+1​+2πn)∪(6π1​+2πn,−25π−25π2+4​​+2πn)∪(5π1​+2πn,−2π+4π2+1​+2πn)∪(4π1​+2πn,−23π−9π2+4​​+2πn)∪(3π1​+2πn,−π+π2+1​+2πn)∪(2π1​+2πn,−2π−π2+4​​+2πn)∪(π1​+2πn,1+2πn)∪(π+2πn,−2−π−π2+4​​+2πn)
Decimal
2πn<x<0.03532…+2πnor0.03536…+2πn<x<0.03972…+2πnor0.03978…+2πn<x<0.04537…+2πnor0.04547…+2πn<x<0.05290…+2πnor0.05305…+2πn<x<0.06340…+2πnor0.06366…+2πn<x<0.07907…+2πnor0.07957…+2πn<x<0.10493…+2πnor0.10610…+2πn<x<0.15531…+2πnor0.15915…+2πn<x<0.29129…+2πnor0.31830…+2πn<x<1+2πnor3.14159…+2πn<x<3.43289…+2πn
Solution steps
tan(x)1​>cot(x1​)
Move cot(x1​)to the left side
tan(x)1​>cot(x1​)
Subtract cot(x1​) from both sidestan(x)1​−cot(x1​)>cot(x1​)−cot(x1​)
tan(x)1​−cot(x1​)>0
tan(x)1​−cot(x1​)>0
Periodicity of tan(x)1​−cot(x1​):Not periodic
The function tan(x)1​−cot(x1​)is not periodic=Notperiodic
Express with sin, cos
tan(x)1​−cot(x1​)>0
Use the basic trigonometric identity: tan(x)=cos(x)sin(x)​cos(x)sin(x)​1​−cot(x1​)>0
Use the basic trigonometric identity: cot(x)=sin(x)cos(x)​cos(x)sin(x)​1​−sin(x1​)cos(x1​)​>0
cos(x)sin(x)​1​−sin(x1​)cos(x1​)​>0
Simplify cos(x)sin(x)​1​−sin(x1​)cos(x1​)​:sin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​
cos(x)sin(x)​1​−sin(x1​)cos(x1​)​
Apply the fraction rule: cb​1​=bc​=sin(x)cos(x)​−sin(x1​)cos(x1​)​
Least Common Multiplier of sin(x),sin(x1​):sin(x)sin(x1​)
sin(x),sin(x1​)
Lowest Common Multiplier (LCM)
Compute an expression comprised of factors that appear either in sin(x) or sin(x1​)=sin(x)sin(x1​)
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 sin(x)sin(x1​)
For sin(x)cos(x)​:multiply the denominator and numerator by sin(x1​)sin(x)cos(x)​=sin(x)sin(x1​)cos(x)sin(x1​)​
For sin(x1​)cos(x1​)​:multiply the denominator and numerator by sin(x)sin(x1​)cos(x1​)​=sin(x1​)sin(x)cos(x1​)sin(x)​
=sin(x)sin(x1​)cos(x)sin(x1​)​−sin(x1​)sin(x)cos(x1​)sin(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=sin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​
sin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​>0
Find the zeroes and undifined points of sin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​for 0≤x<2π
To find the zeroes, set the inequality to zerosin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​=0
sin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​=0,0≤x<2π:x=1,x=−2π−π2+4​​,x=−π+π2+1​,x=−23π−9π2+4​​,x=−2π+4π2+1​,x=−25π−25π2+4​​,x=−3π+9π2+1​,x=−27π−49π2+4​​,x=−4π+16π2+1​,x=−29π−81π2+4​​,x=−2−π−π2+4​​
sin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​=0,0≤x<2π
g(x)f(x)​=0⇒f(x)=0cos(x)sin(x1​)−cos(x1​)sin(x)=0
Rewrite using trig identities
cos(x)sin(x1​)−cos(x1​)sin(x)
Use the Angle Difference identity: sin(s)cos(t)−cos(s)sin(t)=sin(s−t)=sin(x1​−x)
sin(x1​−x)=0
General solutions for sin(x1​−x)=0
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x1​−x=0+2πn,x1​−x=π+2πn
x1​−x=0+2πn,x1​−x=π+2πn
Solve x1​−x=0+2πn:x=−πn−π2n2+1​,x=−πn+π2n2+1​
x1​−x=0+2πn
Multiply both sides by x
x1​−x=0+2πn
Multiply both sides by xx1​x−xx=0⋅x+2πnx
Simplify
x1​x−xx=0⋅x+2πnx
Simplify x1​x:1
x1​x
Multiply fractions: a⋅cb​=ca⋅b​=x1⋅x​
Cancel the common factor: x=1
Simplify −xx:−x2
−xx
Apply exponent rule: ab⋅ac=ab+cxx=x1+1=−x1+1
Add the numbers: 1+1=2=−x2
Simplify 0⋅x:0
0⋅x
Apply rule 0⋅a=0=0
1−x2=0+2πnx
Simplify 0+2πnx:2πnx
0+2πnx
0+2πnx=2πnx=2πnx
1−x2=2πnx
1−x2=2πnx
1−x2=2πnx
Solve 1−x2=2πnx:x=−πn−π2n2+1​,x=−πn+π2n2+1​
1−x2=2πnx
Move 2πnxto the left side
1−x2=2πnx
Subtract 2πnx from both sides1−x2−2πnx=2πnx−2πnx
Simplify1−x2−2πnx=0
1−x2−2πnx=0
Write in the standard form ax2+bx+c=0−x2−2πnx+1=0
Solve with the quadratic formula
−x2−2πnx+1=0
Quadratic Equation Formula:
For a=−1,b=−2πn,c=1x1,2​=2(−1)−(−2πn)±(−2πn)2−4(−1)⋅1​​
x1,2​=2(−1)−(−2πn)±(−2πn)2−4(−1)⋅1​​
Simplify (−2πn)2−4(−1)⋅1​:2π2n2+1​
(−2πn)2−4(−1)⋅1​
Apply rule −(−a)=a=(−2πn)2+4⋅1⋅1​
(−2πn)2=22π2n2
(−2πn)2
Apply exponent rule: (−a)n=an,if n is even(−2πn)2=(2πn)2=(2πn)2
Apply exponent rule: (a⋅b)n=anbn=22π2n2
4⋅1⋅1=4
4⋅1⋅1
Multiply the numbers: 4⋅1⋅1=4=4
=22π2n2+4​
Factor 22π2n2+4:4(π2n2+1)
22π2n2+4
Rewrite as=4π2n2+4⋅1
Factor out common term 4=4(π2n2+1)
=4(π2n2+1)​
Apply radical rule: assuming a≥0,b≥0=4​π2n2+1​
4​=2
4​
Factor the number: 4=22=22​
Apply radical rule: 22​=2=2
=2π2n2+1​
x1,2​=2(−1)−(−2πn)±2π2n2+1​​
Separate the solutionsx1​=2(−1)−(−2πn)+2π2n2+1​​,x2​=2(−1)−(−2πn)−2π2n2+1​​
x=2(−1)−(−2πn)+2π2n2+1​​:−πn−π2n2+1​
2(−1)−(−2πn)+2π2n2+1​​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅12πn+2π2n2+1​​
Multiply the numbers: 2⋅1=2=−22πn+2π2n2+1​​
Apply the fraction rule: −ba​=−ba​=−22πn+2π2n2+1​​
Cancel 22πn+2π2n2+1​​:πn+π2n2+1​
22πn+2π2n2+1​​
Factor out common term 2=22(πn+1+n2π2​)​
Divide the numbers: 22​=1=πn+π2n2+1​
=−(πn+π2n2+1​)
Distribute parentheses=−(πn)−(π2n2+1​)
Apply minus-plus rules+(−a)=−a=−πn−π2n2+1​
x=2(−1)−(−2πn)−2π2n2+1​​:−πn+π2n2+1​
2(−1)−(−2πn)−2π2n2+1​​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅12πn−2π2n2+1​​
Multiply the numbers: 2⋅1=2=−22πn−2π2n2+1​​
Apply the fraction rule: −ba​=−ba​=−22πn−2π2n2+1​​
Cancel 22πn−2π2n2+1​​:πn−π2n2+1​
22πn−2π2n2+1​​
Factor out common term 2=22(πn−1+n2π2​)​
Divide the numbers: 22​=1=πn−π2n2+1​
=−(πn−π2n2+1​)
Distribute parentheses=−(πn)−(−π2n2+1​)
Apply minus-plus rules−(−a)=a,−(a)=−a=−πn+π2n2+1​
The solutions to the quadratic equation are:x=−πn−π2n2+1​,x=−πn+π2n2+1​
x=−πn−π2n2+1​,x=−πn+π2n2+1​
Solve x1​−x=π+2πn:x=−22πn+π+(−2πn−π)2+4​​,x=−22πn+π−(−2πn−π)2+4​​
x1​−x=π+2πn
Multiply both sides by x
x1​−x=π+2πn
Multiply both sides by xx1​x−xx=πx+2πnx
Simplify
x1​x−xx=πx+2πnx
Simplify x1​x:1
x1​x
Multiply fractions: a⋅cb​=ca⋅b​=x1⋅x​
Cancel the common factor: x=1
Simplify −xx:−x2
−xx
Apply exponent rule: ab⋅ac=ab+cxx=x1+1=−x1+1
Add the numbers: 1+1=2=−x2
1−x2=πx+2πnx
1−x2=πx+2πnx
1−x2=πx+2πnx
Solve 1−x2=πx+2πnx:x=−22πn+π+(−2πn−π)2+4​​,x=−22πn+π−(−2πn−π)2+4​​
1−x2=πx+2πnx
Move 2πnxto the left side
1−x2=πx+2πnx
Subtract 2πnx from both sides1−x2−2πnx=πx+2πnx−2πnx
Simplify1−x2−2πnx=πx
1−x2−2πnx=πx
Move πxto the left side
1−x2−2πnx=πx
Subtract πx from both sides1−x2−2πnx−πx=πx−πx
Simplify1−x2−2πnx−πx=0
1−x2−2πnx−πx=0
Write in the standard form ax2+bx+c=0−x2−(2πn+π)x+1=0
Solve with the quadratic formula
−x2−(2πn+π)x+1=0
Quadratic Equation Formula:
For a=−1,b=−2πn−π,c=1x1,2​=2(−1)−(−2πn−π)±(−2πn−π)2−4(−1)⋅1​​
x1,2​=2(−1)−(−2πn−π)±(−2πn−π)2−4(−1)⋅1​​
Simplify (−2πn−π)2−4(−1)⋅1​:(−2πn−π)2+4​
(−2πn−π)2−4(−1)⋅1​
Apply rule −(−a)=a=(−2πn−π)2+4⋅1⋅1​
Multiply the numbers: 4⋅1⋅1=4=(−2πn−π)2+4​
x1,2​=2(−1)−(−2πn−π)±(−2πn−π)2+4​​
Separate the solutionsx1​=2(−1)−(−2πn−π)+(−2πn−π)2+4​​,x2​=2(−1)−(−2πn−π)−(−2πn−π)2+4​​
x=2(−1)−(−2πn−π)+(−2πn−π)2+4​​:−22πn+π+(−2πn−π)2+4​​
2(−1)−(−2πn−π)+(−2πn−π)2+4​​
Remove parentheses: (−a)=−a=−2⋅1−(−2πn−π)+(−2πn−π)2+4​​
Multiply the numbers: 2⋅1=2=−2−(−2πn−π)+(−2πn−π)2+4​​
Apply the fraction rule: −ba​=−ba​=−2−(−2πn−π)+(−2πn−π)2+4​​
−(−2πn−π):2πn+π
−(−2πn−π)
Distribute parentheses=−(−2πn)−(−π)
Apply minus-plus rules−(−a)=a=2πn+π
=−22πn+π+(−2πn−π)2+4​​
x=2(−1)−(−2πn−π)−(−2πn−π)2+4​​:−22πn+π−(−2πn−π)2+4​​
2(−1)−(−2πn−π)−(−2πn−π)2+4​​
Remove parentheses: (−a)=−a=−2⋅1−(−2πn−π)−(−2πn−π)2+4​​
Multiply the numbers: 2⋅1=2=−2−(−2πn−π)−(−2πn−π)2+4​​
Apply the fraction rule: −ba​=−ba​=−2−(−2πn−π)−(−2πn−π)2+4​​
−(−2πn−π):2πn+π
−(−2πn−π)
Distribute parentheses=−(−2πn)−(−π)
Apply minus-plus rules−(−a)=a=2πn+π
=−22πn+π−(−2πn−π)2+4​​
The solutions to the quadratic equation are:x=−22πn+π+(−2πn−π)2+4​​,x=−22πn+π−(−2πn−π)2+4​​
x=−22πn+π+(−2πn−π)2+4​​,x=−22πn+π−(−2πn−π)2+4​​
x=−πn−π2n2+1​,x=−πn+π2n2+1​,x=−22πn+π+(−2πn−π)2+4​​,x=−22πn+π−(−2πn−π)2+4​​
Solutions for the range 0≤x<2πx=1,x=−2π−π2+4​​,x=−π+π2+1​,x=−23π−9π2+4​​,x=−2π+4π2+1​,x=−25π−25π2+4​​,x=−3π+9π2+1​,x=−27π−49π2+4​​,x=−4π+16π2+1​,x=−29π−81π2+4​​,x=−2−π−π2+4​​
Find the undefined points:x=π,x=π1​,x=2π1​,x=3π1​,x=4π1​,x=5π1​,x=6π1​,x=7π1​,x=8π1​,x=9π1​
Find the zeros of the denominatorsin(x)sin(x1​)=0
Solving each part separatelysin(x)=0orsin(x1​)=0
sin(x)=0,0≤x<2π:x=0,x=π
sin(x)=0,0≤x<2π
General solutions for sin(x)=0
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=0+2πn,x=π+2πn
x=0+2πn,x=π+2πn
Solve x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn,x=π+2πn
Solutions for the range 0≤x<2πx=0,x=π
sin(x1​)=0,0≤x<2π:x=π1​,x=2π1​,x=3π1​,x=4π1​,x=5π1​,x=6π1​,x=7π1​,x=8π1​,x=9π1​
sin(x1​)=0,0≤x<2π
General solutions for sin(x1​)=0
sin(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x1​=0+2πn,x1​=π+2πn
x1​=0+2πn,x1​=π+2πn
Solve x1​=0+2πn:x=2πn1​;n=0
x1​=0+2πn
Multiply both sides by x
x1​=0+2πn
Multiply both sides by xx1​x=0⋅x+2πnx
Simplify
1=0+2πnx
Simplify 0+2πnx:2πnx
0+2πnx
0+2πnx=2πnx=2πnx
1=2πnx
1=2πnx
Switch sides2πnx=1
Divide both sides by 2πn;n=0
2πnx=1
Divide both sides by 2πn;n=02πn2πnx​=2πn1​;n=0
Simplifyx=2πn1​;n=0
x=2πn1​;n=0
Solve x1​=π+2πn:x=π(1+2n)1​;n=−21​
x1​=π+2πn
Multiply both sides by x
x1​=π+2πn
Multiply both sides by xx1​x=πx+2πnx
Simplify1=πx+2πnx
1=πx+2πnx
Switch sidesπx+2πnx=1
Factor πx+2πnx:πx(1+2n)
πx+2πnx
Factor out common term xπ=xπ(1+2n)
πx(1+2n)=1
Divide both sides by π(1+2n);n=−21​
πx(1+2n)=1
Divide both sides by π(1+2n);n=−21​π(1+2n)πx(1+2n)​=π(1+2n)1​;n=−21​
Simplifyx=π(1+2n)1​;n=−21​
x=π(1+2n)1​;n=−21​
x=2πn1​,x=π(1+2n)1​;n=0,n=−21​
Solutions for the range 0≤x<2πx=π1​,x=2π1​,x=3π1​,x=4π1​,x=5π1​,x=6π1​,x=7π1​,x=8π1​,x=9π1​
Combine all the solutionsx=0,x=π,x=π1​,x=2π1​,x=3π1​,x=4π1​,x=5π1​,x=6π1​,x=7π1​,x=8π1​,x=9π1​
Since the equation is undefined for:0x=π,x=π1​,x=2π1​,x=3π1​,x=4π1​,x=5π1​,x=6π1​,x=7π1​,x=8π1​,x=9π1​
−29π−81π2+4​​,9π1​,−4π+16π2+1​,8π1​,−27π−49π2+4​​,7π1​,−3π+9π2+1​,6π1​,−25π−25π2+4​​,5π1​,−2π+4π2+1​,4π1​,−23π−9π2+4​​,3π1​,−π+π2+1​,2π1​,−2π−π2+4​​,π1​,1,π,−2−π−π2+4​​
Identify the intervals0<x<−29π−81π2+4​​,−29π−81π2+4​​<x<9π1​,9π1​<x<−4π+16π2+1​,−4π+16π2+1​<x<8π1​,8π1​<x<−27π−49π2+4​​,−27π−49π2+4​​<x<7π1​,7π1​<x<−3π+9π2+1​,−3π+9π2+1​<x<6π1​,6π1​<x<−25π−25π2+4​​,−25π−25π2+4​​<x<5π1​,5π1​<x<−2π+4π2+1​,−2π+4π2+1​<x<4π1​,4π1​<x<−23π−9π2+4​​,−23π−9π2+4​​<x<3π1​,3π1​<x<−π+π2+1​,−π+π2+1​<x<2π1​,2π1​<x<−2π−π2+4​​,−2π−π2+4​​<x<π1​,π1​<x<1,1<x<π,π<x<−2−π−π2+4​​,−2−π−π2+4​​<x<2π
Summarize in a table:cos(x)sin(x1​)−cos(x1​)sin(x)sin(x)sin(x1​)sin(x)sin(x1​)cos(x)sin(x1​)−cos(x1​)sin(x)​​x=0Undefined0UndefinedUndefined​0<x<−29π−81π2+4​​++++​x=−29π−81π2+4​​0+−0​−29π−81π2+4​​<x<9π1​++−−​x=9π1​++0Undefined​9π1​<x<−4π+16π2+1​++++​x=−4π+16π2+1​0++0​−4π+16π2+1​<x<8π1​−++−​x=8π1​−+0Undefined​8π1​<x<−27π−49π2+4​​−+−+​x=−27π−49π2+4​​0+−0​−27π−49π2+4​​<x<7π1​++−−​x=7π1​++0Undefined​7π1​<x<−3π+9π2+1​++++​x=−3π+9π2+1​0++0​−3π+9π2+1​<x<6π1​−++−​x=6π1​−+0Undefined​6π1​<x<−25π−25π2+4​​−+−+​x=−25π−25π2+4​​0+−0​−25π−25π2+4​​<x<5π1​++−−​x=5π1​++0Undefined​5π1​<x<−2π+4π2+1​++++​x=−2π+4π2+1​0++0​−2π+4π2+1​<x<4π1​−++−​x=4π1​−+0Undefined​4π1​<x<−23π−9π2+4​​−+−+​x=−23π−9π2+4​​0+−0​−23π−9π2+4​​<x<3π1​++−−​x=3π1​++0Undefined​3π1​<x<−π+π2+1​++++​x=−π+π2+1​0++0​−π+π2+1​<x<2π1​−++−​x=2π1​−+0Undefined​2π1​<x<−2π−π2+4​​−+−+​x=−2π−π2+4​​0+−0​−2π−π2+4​​<x<π1​++−−​x=π1​++0Undefined​π1​<x<1++++​x=10++0​1<x<π−++−​x=π−0+Undefined​π<x<−2−π−π2+4​​−−++​x=−2−π−π2+4​​0−+0​−2−π−π2+4​​<x<2π+−+−​x=2π+0+Undefined​​
Identify the intervals that satisfy the required condition: >00<x<−29π−81π2+4​​or9π1​<x<−4π+16π2+1​or8π1​<x<−27π−49π2+4​​or7π1​<x<−3π+9π2+1​or6π1​<x<−25π−25π2+4​​or5π1​<x<−2π+4π2+1​or4π1​<x<−23π−9π2+4​​or3π1​<x<−π+π2+1​or2π1​<x<−2π−π2+4​​orπ1​<x<1orπ<x<−2−π−π2+4​​
Apply the periodicity of tan(x)1​−cot(x1​)2πn<x<−29π−81π2+4​​+2πnor9π1​+2πn<x<−4π+16π2+1​+2πnor8π1​+2πn<x<−27π−49π2+4​​+2πnor7π1​+2πn<x<−3π+9π2+1​+2πnor6π1​+2πn<x<−25π−25π2+4​​+2πnor5π1​+2πn<x<−2π+4π2+1​+2πnor4π1​+2πn<x<−23π−9π2+4​​+2πnor3π1​+2πn<x<−π+π2+1​+2πnor2π1​+2πn<x<−2π−π2+4​​+2πnorπ1​+2πn<x<1+2πnorπ+2πn<x<−2−π−π2+4​​+2πn

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