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Popular Trigonometry >

tan(3x)=5tan(x)

  • Pre Algebra
  • Algebra
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Solution

tan(3x)=5tan(x)

Solution

x=πn,x=−0.36136…+πn,x=0.36136…+πn
+1
Degrees
x=0∘+180∘n,x=−20.70481…∘+180∘n,x=20.70481…∘+180∘n
Solution steps
tan(3x)=5tan(x)
Subtract 5tan(x) from both sidestan(3x)−5tan(x)=0
Rewrite using trig identities
tan(3x)−5tan(x)
tan(3x)=1−3tan2(x)3tan(x)−tan3(x)​
tan(3x)
Rewrite using trig identities
tan(3x)
Rewrite as=tan(2x+x)
Use the Angle Sum identity: tan(s+t)=1−tan(s)tan(t)tan(s)+tan(t)​=1−tan(2x)tan(x)tan(2x)+tan(x)​
=1−tan(2x)tan(x)tan(2x)+tan(x)​
Use the Double Angle identity: tan(2x)=1−tan2(x)2tan(x)​=1−1−tan2(x)2tan(x)​tan(x)1−tan2(x)2tan(x)​+tan(x)​
Simplify 1−1−tan2(x)2tan(x)​tan(x)1−tan2(x)2tan(x)​+tan(x)​:1−3tan2(x)3tan(x)−tan3(x)​
1−1−tan2(x)2tan(x)​tan(x)1−tan2(x)2tan(x)​+tan(x)​
1−tan2(x)2tan(x)​tan(x)=1−tan2(x)2tan2(x)​
1−tan2(x)2tan(x)​tan(x)
Multiply fractions: a⋅cb​=ca⋅b​=1−tan2(x)2tan(x)tan(x)​
2tan(x)tan(x)=2tan2(x)
2tan(x)tan(x)
Apply exponent rule: ab⋅ac=ab+ctan(x)tan(x)=tan1+1(x)=2tan1+1(x)
Add the numbers: 1+1=2=2tan2(x)
=1−tan2(x)2tan2(x)​
=1−−tan2(x)+12tan2(x)​−tan2(x)+12tan(x)​+tan(x)​
Join 1−tan2(x)2tan(x)​+tan(x):1−tan2(x)3tan(x)−tan3(x)​
1−tan2(x)2tan(x)​+tan(x)
Convert element to fraction: tan(x)=1−tan2(x)tan(x)(1−tan2(x))​=1−tan2(x)2tan(x)​+1−tan2(x)tan(x)(1−tan2(x))​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1−tan2(x)2tan(x)+tan(x)(1−tan2(x))​
Expand 2tan(x)+tan(x)(1−tan2(x)):3tan(x)−tan3(x)
2tan(x)+tan(x)(1−tan2(x))
Expand tan(x)(1−tan2(x)):tan(x)−tan3(x)
tan(x)(1−tan2(x))
Apply the distributive law: a(b−c)=ab−aca=tan(x),b=1,c=tan2(x)=tan(x)1−tan(x)tan2(x)
=1tan(x)−tan2(x)tan(x)
Simplify 1⋅tan(x)−tan2(x)tan(x):tan(x)−tan3(x)
1tan(x)−tan2(x)tan(x)
1⋅tan(x)=tan(x)
1tan(x)
Multiply: 1⋅tan(x)=tan(x)=tan(x)
tan2(x)tan(x)=tan3(x)
tan2(x)tan(x)
Apply exponent rule: ab⋅ac=ab+ctan2(x)tan(x)=tan2+1(x)=tan2+1(x)
Add the numbers: 2+1=3=tan3(x)
=tan(x)−tan3(x)
=tan(x)−tan3(x)
=2tan(x)+tan(x)−tan3(x)
Add similar elements: 2tan(x)+tan(x)=3tan(x)=3tan(x)−tan3(x)
=1−tan2(x)3tan(x)−tan3(x)​
=1−−tan2(x)+12tan2(x)​1−tan2(x)3tan(x)−tan3(x)​​
Apply the fraction rule: acb​​=c⋅ab​=(1−tan2(x))(1−1−tan2(x)2tan2(x)​)3tan(x)−tan3(x)​
Join 1−1−tan2(x)2tan2(x)​:1−tan2(x)1−3tan2(x)​
1−1−tan2(x)2tan2(x)​
Convert element to fraction: 1=1−tan2(x)1(1−tan2(x))​=1−tan2(x)1(1−tan2(x))​−1−tan2(x)2tan2(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1−tan2(x)1(1−tan2(x))−2tan2(x)​
1⋅(1−tan2(x))−2tan2(x)=1−3tan2(x)
1(1−tan2(x))−2tan2(x)
1⋅(1−tan2(x))=1−tan2(x)
1(1−tan2(x))
Multiply: 1⋅(1−tan2(x))=(1−tan2(x))=1−tan2(x)
Remove parentheses: (a)=a=1−tan2(x)
=1−tan2(x)−2tan2(x)
Add similar elements: −tan2(x)−2tan2(x)=−3tan2(x)=1−3tan2(x)
=1−tan2(x)1−3tan2(x)​
=−tan2(x)+1−3tan2(x)+1​(−tan2(x)+1)3tan(x)−tan3(x)​
Multiply (1−tan2(x))1−tan2(x)1−3tan2(x)​:1−3tan2(x)
(1−tan2(x))1−tan2(x)1−3tan2(x)​
Multiply fractions: a⋅cb​=ca⋅b​=1−tan2(x)(1−3tan2(x))(1−tan2(x))​
Cancel the common factor: 1−tan2(x)=1−3tan2(x)
=1−3tan2(x)3tan(x)−tan3(x)​
=1−3tan2(x)3tan(x)−tan3(x)​
=1−3tan2(x)3tan(x)−tan3(x)​−5tan(x)
Simplify 1−3tan2(x)3tan(x)−tan3(x)​−5tan(x):1−3tan2(x)−2tan(x)+14tan3(x)​
1−3tan2(x)3tan(x)−tan3(x)​−5tan(x)
Convert element to fraction: 5tan(x)=1−3tan2(x)5tan(x)(1−3tan2(x))​=1−3tan2(x)3tan(x)−tan3(x)​−1−3tan2(x)5tan(x)(1−3tan2(x))​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1−3tan2(x)3tan(x)−tan3(x)−5tan(x)(1−3tan2(x))​
Expand 3tan(x)−tan3(x)−5tan(x)(1−3tan2(x)):−2tan(x)+14tan3(x)
3tan(x)−tan3(x)−5tan(x)(1−3tan2(x))
Expand −5tan(x)(1−3tan2(x)):−5tan(x)+15tan3(x)
−5tan(x)(1−3tan2(x))
Apply the distributive law: a(b−c)=ab−aca=−5tan(x),b=1,c=3tan2(x)=−5tan(x)⋅1−(−5tan(x))⋅3tan2(x)
Apply minus-plus rules−(−a)=a=−5⋅1⋅tan(x)+5⋅3tan2(x)tan(x)
Simplify −5⋅1⋅tan(x)+5⋅3tan2(x)tan(x):−5tan(x)+15tan3(x)
−5⋅1⋅tan(x)+5⋅3tan2(x)tan(x)
5⋅1⋅tan(x)=5tan(x)
5⋅1⋅tan(x)
Multiply the numbers: 5⋅1=5=5tan(x)
5⋅3tan2(x)tan(x)=15tan3(x)
5⋅3tan2(x)tan(x)
Multiply the numbers: 5⋅3=15=15tan2(x)tan(x)
Apply exponent rule: ab⋅ac=ab+ctan2(x)tan(x)=tan2+1(x)=15tan2+1(x)
Add the numbers: 2+1=3=15tan3(x)
=−5tan(x)+15tan3(x)
=−5tan(x)+15tan3(x)
=3tan(x)−tan3(x)−5tan(x)+15tan3(x)
Simplify 3tan(x)−tan3(x)−5tan(x)+15tan3(x):−2tan(x)+14tan3(x)
3tan(x)−tan3(x)−5tan(x)+15tan3(x)
Add similar elements: −tan3(x)+15tan3(x)=14tan3(x)=3tan(x)+14tan3(x)−5tan(x)
Add similar elements: 3tan(x)−5tan(x)=−2tan(x)=−2tan(x)+14tan3(x)
=−2tan(x)+14tan3(x)
=1−3tan2(x)−2tan(x)+14tan3(x)​
=1−3tan2(x)−2tan(x)+14tan3(x)​
1−3tan2(x)14tan3(x)−2tan(x)​=0
Solve by substitution
1−3tan2(x)14tan3(x)−2tan(x)​=0
Let: tan(x)=u1−3u214u3−2u​=0
1−3u214u3−2u​=0:u=0,u=−77​​,u=77​​
1−3u214u3−2u​=0
g(x)f(x)​=0⇒f(x)=014u3−2u=0
Solve 14u3−2u=0:u=0,u=−77​​,u=77​​
14u3−2u=0
Factor 14u3−2u:2u(7​u+1)(7​u−1)
14u3−2u
Factor out common term 2u:2u(7u2−1)
14u3−2u
Apply exponent rule: ab+c=abacu3=u2u=14u2u−2u
Rewrite 14 as 2⋅7=2⋅7u2u−2u
Factor out common term 2u=2u(7u2−1)
=2u(7u2−1)
Factor 7u2−1:(7​u+1)(7​u−1)
7u2−1
Rewrite 7u2−1 as (7​u)2−12
7u2−1
Apply radical rule: a=(a​)27=(7​)2=(7​)2u2−1
Rewrite 1 as 12=(7​)2u2−12
Apply exponent rule: ambm=(ab)m(7​)2u2=(7​u)2=(7​u)2−12
=(7​u)2−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)(7​u)2−12=(7​u+1)(7​u−1)=(7​u+1)(7​u−1)
=2u(7​u+1)(7​u−1)
2u(7​u+1)(7​u−1)=0
Using the Zero Factor Principle: If ab=0then a=0or b=0u=0or7​u+1=0or7​u−1=0
Solve 7​u+1=0:u=−77​​
7​u+1=0
Move 1to the right side
7​u+1=0
Subtract 1 from both sides7​u+1−1=0−1
Simplify7​u=−1
7​u=−1
Divide both sides by 7​
7​u=−1
Divide both sides by 7​7​7​u​=7​−1​
Simplify
7​7​u​=7​−1​
Simplify 7​7​u​:u
7​7​u​
Cancel the common factor: 7​=u
Simplify 7​−1​:−77​​
7​−1​
Apply the fraction rule: b−a​=−ba​=−7​1​
Rationalize −7​1​:−77​​
−7​1​
Multiply by the conjugate 7​7​​=−7​7​1⋅7​​
1⋅7​=7​
7​7​=7
7​7​
Apply radical rule: a​a​=a7​7​=7=7
=−77​​
=−77​​
u=−77​​
u=−77​​
u=−77​​
Solve 7​u−1=0:u=77​​
7​u−1=0
Move 1to the right side
7​u−1=0
Add 1 to both sides7​u−1+1=0+1
Simplify7​u=1
7​u=1
Divide both sides by 7​
7​u=1
Divide both sides by 7​7​7​u​=7​1​
Simplify
7​7​u​=7​1​
Simplify 7​7​u​:u
7​7​u​
Cancel the common factor: 7​=u
Simplify 7​1​:77​​
7​1​
Multiply by the conjugate 7​7​​=7​7​1⋅7​​
1⋅7​=7​
7​7​=7
7​7​
Apply radical rule: a​a​=a7​7​=7=7
=77​​
u=77​​
u=77​​
u=77​​
The solutions areu=0,u=−77​​,u=77​​
u=0,u=−77​​,u=77​​
Verify Solutions
Find undefined (singularity) points:u=3​1​,u=−3​1​
Take the denominator(s) of 1−3u214u3−2u​ and compare to zero
Solve 1−3u2=0:u=3​1​,u=−3​1​
1−3u2=0
Move 1to the right side
1−3u2=0
Subtract 1 from both sides1−3u2−1=0−1
Simplify−3u2=−1
−3u2=−1
Divide both sides by −3
−3u2=−1
Divide both sides by −3−3−3u2​=−3−1​
Simplifyu2=31​
u2=31​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=31​​,u=−31​​
31​​=3​1​
31​​
Apply radical rule: ba​​=b​a​​,a≥0,b≥0=3​1​​
Apply radical rule: 1​=11​=1=3​1​
−31​​=−3​1​
−31​​
Apply radical rule: ba​​=b​a​​,a≥0,b≥0=−3​1​​
Apply radical rule: 1​=11​=1=−3​1​
u=3​1​,u=−3​1​
The following points are undefinedu=3​1​,u=−3​1​
Combine undefined points with solutions:
u=0,u=−77​​,u=77​​
Substitute back u=tan(x)tan(x)=0,tan(x)=−77​​,tan(x)=77​​
tan(x)=0,tan(x)=−77​​,tan(x)=77​​
tan(x)=0:x=πn
tan(x)=0
General solutions for tan(x)=0
tan(x) periodicity table with πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​tan(x)033​​13​±∞−3​−1−33​​​​
x=0+πn
x=0+πn
Solve x=0+πn:x=πn
x=0+πn
0+πn=πnx=πn
x=πn
tan(x)=−77​​:x=arctan(−77​​)+πn
tan(x)=−77​​
Apply trig inverse properties
tan(x)=−77​​
General solutions for tan(x)=−77​​tan(x)=−a⇒x=arctan(−a)+πnx=arctan(−77​​)+πn
x=arctan(−77​​)+πn
tan(x)=77​​:x=arctan(77​​)+πn
tan(x)=77​​
Apply trig inverse properties
tan(x)=77​​
General solutions for tan(x)=77​​tan(x)=a⇒x=arctan(a)+πnx=arctan(77​​)+πn
x=arctan(77​​)+πn
Combine all the solutionsx=πn,x=arctan(−77​​)+πn,x=arctan(77​​)+πn
Show solutions in decimal formx=πn,x=−0.36136…+πn,x=0.36136…+πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for tan(3x)=5tan(x) ?

    The general solution for tan(3x)=5tan(x) is x=pin,x=-0.36136…+pin,x=0.36136…+pin
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