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Popular Trigonometry >

3tan(3x)=tan(x)

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Solution

3tan(3x)=tan(x)

Solution

x=πn
+1
Degrees
x=0∘+180∘n
Solution steps
3tan(3x)=tan(x)
Subtract tan(x) from both sides3tan(3x)−tan(x)=0
Rewrite using trig identities
−tan(x)+3tan(3x)
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)​
=−tan(x)+3⋅1−3tan2(x)3tan(x)−tan3(x)​
Simplify −tan(x)+3⋅1−3tan2(x)3tan(x)−tan3(x)​:1−3tan2(x)8tan(x)​
−tan(x)+3⋅1−3tan2(x)3tan(x)−tan3(x)​
Multiply 3⋅1−3tan2(x)3tan(x)−tan3(x)​:1−3tan2(x)3(3tan(x)−tan3(x))​
3⋅1−3tan2(x)3tan(x)−tan3(x)​
Multiply fractions: a⋅cb​=ca⋅b​=1−3tan2(x)(3tan(x)−tan3(x))⋅3​
=−tan(x)+−3tan2(x)+13(3tan(x)−tan3(x))​
Convert element to fraction: tan(x)=1−3tan2(x)tan(x)(1−3tan2(x))​=1−3tan2(x)(3tan(x)−tan3(x))⋅3​−1−3tan2(x)tan(x)(1−3tan2(x))​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1−3tan2(x)(3tan(x)−tan3(x))⋅3−tan(x)(1−3tan2(x))​
Expand (3tan(x)−tan3(x))⋅3−tan(x)(1−3tan2(x)):8tan(x)
(3tan(x)−tan3(x))⋅3−tan(x)(1−3tan2(x))
=3(3tan(x)−tan3(x))−tan(x)(1−3tan2(x))
Expand 3(3tan(x)−tan3(x)):9tan(x)−3tan3(x)
3(3tan(x)−tan3(x))
Apply the distributive law: a(b−c)=ab−aca=3,b=3tan(x),c=tan3(x)=3⋅3tan(x)−3tan3(x)
Multiply the numbers: 3⋅3=9=9tan(x)−3tan3(x)
=9tan(x)−3tan3(x)−tan(x)(1−3tan2(x))
Expand −tan(x)(1−3tan2(x)):−tan(x)+3tan3(x)
−tan(x)(1−3tan2(x))
Apply the distributive law: a(b−c)=ab−aca=−tan(x),b=1,c=3tan2(x)=−tan(x)⋅1−(−tan(x))⋅3tan2(x)
Apply minus-plus rules−(−a)=a=−1⋅tan(x)+3tan2(x)tan(x)
Simplify −1⋅tan(x)+3tan2(x)tan(x):−tan(x)+3tan3(x)
−1⋅tan(x)+3tan2(x)tan(x)
1⋅tan(x)=tan(x)
1⋅tan(x)
Multiply: 1⋅tan(x)=tan(x)=tan(x)
3tan2(x)tan(x)=3tan3(x)
3tan2(x)tan(x)
Apply exponent rule: ab⋅ac=ab+ctan2(x)tan(x)=tan2+1(x)=3tan2+1(x)
Add the numbers: 2+1=3=3tan3(x)
=−tan(x)+3tan3(x)
=−tan(x)+3tan3(x)
=9tan(x)−3tan3(x)−tan(x)+3tan3(x)
Simplify 9tan(x)−3tan3(x)−tan(x)+3tan3(x):8tan(x)
9tan(x)−3tan3(x)−tan(x)+3tan3(x)
Add similar elements: −3tan3(x)+3tan3(x)=0=9tan(x)−tan(x)
Add similar elements: 9tan(x)−tan(x)=8tan(x)=8tan(x)
=8tan(x)
=1−3tan2(x)8tan(x)​
=1−3tan2(x)8tan(x)​
1−3tan2(x)8tan(x)​=0
Solve by substitution
1−3tan2(x)8tan(x)​=0
Let: tan(x)=u1−3u28u​=0
1−3u28u​=0:u=0
1−3u28u​=0
g(x)f(x)​=0⇒f(x)=08u=0
Divide both sides by 8
8u=0
Divide both sides by 888u​=80​
Simplifyu=0
u=0
Verify Solutions
Find undefined (singularity) points:u=3​1​,u=−3​1​
Take the denominator(s) of 1−3u28u​ 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
Substitute back u=tan(x)tan(x)=0
tan(x)=0
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
Combine all the solutionsx=πn

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Popular Examples

sin(x)=(sqrt(5))/5 ,sin(2x)0=1-cos(2pix)tan(θ)= 12/5 ,0<= θ<= pi/2sin(x)=(420)/(2.3)cos(x)=(sqrt(14))/(14)

Frequently Asked Questions (FAQ)

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

    The general solution for 3tan(3x)=tan(x) is x=pin
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