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Popular Trigonometry >

2tan(60-x)=tan(x)

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Solution

2tan(60−x)=tan(x)

Solution

x=−1.46681…+πn,x=0.20575…+πn
+1
Degrees
x=−84.04237…∘+180∘n,x=11.78914…∘+180∘n
Solution steps
2tan(60−x)=tan(x)
Subtract tan(x) from both sides2tan(60−x)−tan(x)=0
Rewrite using trig identities
−tan(x)+2tan(60−x)
Use the Angle Difference identity: tan(s−t)=1+tan(s)tan(t)tan(s)−tan(t)​=−tan(x)+2⋅1+tan(60)tan(x)tan(60)−tan(x)​
Simplify −tan(x)+2⋅1+tan(60)tan(x)tan(60)−tan(x)​:1+tan(60)tan(x)2tan(60)−3tan(x)−tan(60)tan2(x)​
−tan(x)+2⋅1+tan(60)tan(x)tan(60)−tan(x)​
Multiply 2⋅1+tan(60)tan(x)tan(60)−tan(x)​:1+tan(60)tan(x)2(−tan(x)+tan(60))​
2⋅1+tan(60)tan(x)tan(60)−tan(x)​
Multiply fractions: a⋅cb​=ca⋅b​=1+tan(60)tan(x)(tan(60)−tan(x))⋅2​
=−tan(x)+tan(60)tan(x)+12(−tan(x)+tan(60))​
Convert element to fraction: tan(x)=1+tan(60)tan(x)tan(x)(1+tan(60)tan(x))​=1+tan(60)tan(x)(tan(60)−tan(x))⋅2​−1+tan(60)tan(x)tan(x)(1+tan(60)tan(x))​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=1+tan(60)tan(x)(tan(60)−tan(x))⋅2−tan(x)(1+tan(60)tan(x))​
Expand (tan(60)−tan(x))⋅2−tan(x)(1+tan(60)tan(x)):2tan(60)−3tan(x)−tan(60)tan2(x)
(tan(60)−tan(x))⋅2−tan(x)(1+tan(60)tan(x))
=2(tan(60)−tan(x))−tan(x)(1+tan(60)tan(x))
Expand 2(tan(60)−tan(x)):2tan(60)−2tan(x)
2(tan(60)−tan(x))
Apply the distributive law: a(b−c)=ab−aca=2,b=tan(60),c=tan(x)=2tan(60)−2tan(x)
=2tan(60)−2tan(x)−tan(x)(1+tan(60)tan(x))
Expand −tan(x)(1+tan(60)tan(x)):−tan(x)−tan(60)tan2(x)
−tan(x)(1+tan(60)tan(x))
Apply the distributive law: a(b+c)=ab+aca=−tan(x),b=1,c=tan(60)tan(x)=−tan(x)⋅1+(−tan(x))tan(60)tan(x)
Apply minus-plus rules+(−a)=−a=−1⋅tan(x)−tan(60)tan(x)tan(x)
Simplify −1⋅tan(x)−tan(60)tan(x)tan(x):−tan(x)−tan(60)tan2(x)
−1⋅tan(x)−tan(60)tan(x)tan(x)
1⋅tan(x)=tan(x)
1⋅tan(x)
Multiply: 1⋅tan(x)=tan(x)=tan(x)
tan(60)tan(x)tan(x)=tan(60)tan2(x)
tan(60)tan(x)tan(x)
Apply exponent rule: ab⋅ac=ab+ctan(x)tan(x)=tan1+1(x)=tan(60)tan1+1(x)
Add the numbers: 1+1=2=tan(60)tan2(x)
=−tan(x)−tan(60)tan2(x)
=−tan(x)−tan(60)tan2(x)
=2tan(60)−2tan(x)−tan(x)−tan(60)tan2(x)
Add similar elements: −2tan(x)−tan(x)=−3tan(x)=2tan(60)−3tan(x)−tan(60)tan2(x)
=1+tan(60)tan(x)2tan(60)−3tan(x)−tan(60)tan2(x)​
=1+tan(60)tan(x)2tan(60)−3tan(x)−tan(60)tan2(x)​
1+tan(60)tan(x)2tan(60)−3tan(x)−tan(60)tan2(x)​=0
Solve by substitution
1+tan(60)tan(x)2tan(60)−3tan(x)−tan(60)tan2(x)​=0
Let: tan(x)=u1+tan(60)u2tan(60)−3u−tan(60)u2​=0
1+tan(60)u2tan(60)−3u−tan(60)u2​=0:u=−2tan(60)3+9+8tan2(60)​​,u=2tan(60)8tan2(60)+9​−3​
1+tan(60)u2tan(60)−3u−tan(60)u2​=0
g(x)f(x)​=0⇒f(x)=02tan(60)−3u−tan(60)u2=0
Solve 2tan(60)−3u−tan(60)u2=0:u=−2tan(60)3+9+8tan2(60)​​,u=2tan(60)8tan2(60)+9​−3​
2tan(60)−3u−tan(60)u2=0
Write in the standard form ax2+bx+c=0−tan(60)u2−3u+2tan(60)=0
Solve with the quadratic formula
−tan(60)u2−3u+2tan(60)=0
Quadratic Equation Formula:
For a=−tan(60),b=−3,c=2tan(60)u1,2​=2(−tan(60))−(−3)±(−3)2−4(−tan(60))⋅2tan(60)​​
u1,2​=2(−tan(60))−(−3)±(−3)2−4(−tan(60))⋅2tan(60)​​
(−3)2−4(−tan(60))⋅2tan(60)​=9+8tan2(60)​
(−3)2−4(−tan(60))⋅2tan(60)​
Apply rule −(−a)=a=(−3)2+4tan(60)⋅2tan(60)​
(−3)2=32
(−3)2
Apply exponent rule: (−a)n=an,if n is even(−3)2=32=32
4tan(60)⋅2tan(60)=8tan2(60)
4tan(60)⋅2tan(60)
Multiply the numbers: 4⋅2=8=8tan(60)tan(60)
Apply exponent rule: ab⋅ac=ab+ctan(60)tan(60)=tan1+1(60)=8tan1+1(60)
Add the numbers: 1+1=2=8tan2(60)
=32+8tan2(60)​
32=9=9+8tan2(60)​
u1,2​=2(−tan(60))−(−3)±9+8tan2(60)​​
Separate the solutionsu1​=2(−tan(60))−(−3)+9+8tan2(60)​​,u2​=2(−tan(60))−(−3)−9+8tan2(60)​​
u=2(−tan(60))−(−3)+9+8tan2(60)​​:−2tan(60)3+9+8tan2(60)​​
2(−tan(60))−(−3)+9+8tan2(60)​​
Remove parentheses: (−a)=−a,−(−a)=a=−2tan(60)3+9+8tan2(60)​​
Apply the fraction rule: −ba​=−ba​=−2tan(60)3+9+8tan2(60)​​
u=2(−tan(60))−(−3)−9+8tan2(60)​​:2tan(60)8tan2(60)+9​−3​
2(−tan(60))−(−3)−9+8tan2(60)​​
Remove parentheses: (−a)=−a,−(−a)=a=−2tan(60)3−9+8tan2(60)​​
Apply the fraction rule: −b−a​=ba​3−9+8tan2(60)​=−(8tan2(60)+9​−3)=2tan(60)8tan2(60)+9​−3​
The solutions to the quadratic equation are:u=−2tan(60)3+9+8tan2(60)​​,u=2tan(60)8tan2(60)+9​−3​
u=−2tan(60)3+9+8tan2(60)​​,u=2tan(60)8tan2(60)+9​−3​
Verify Solutions
Find undefined (singularity) points:u=−tan(60)1​
Take the denominator(s) of 1+tan(60)u2tan(60)−3u−tan(60)u2​ and compare to zero
Solve 1+tan(60)u=0:u=−tan(60)1​
1+tan(60)u=0
Move 1to the right side
1+tan(60)u=0
Subtract 1 from both sides1+tan(60)u−1=0−1
Simplifytan(60)u=−1
tan(60)u=−1
Divide both sides by tan(60)
tan(60)u=−1
Divide both sides by tan(60)tan(60)tan(60)u​=tan(60)−1​
Simplifyu=−tan(60)1​
u=−tan(60)1​
The following points are undefinedu=−tan(60)1​
Combine undefined points with solutions:
u=−2tan(60)3+9+8tan2(60)​​,u=2tan(60)8tan2(60)+9​−3​
Substitute back u=tan(x)tan(x)=−2tan(60)3+9+8tan2(60)​​,tan(x)=2tan(60)8tan2(60)+9​−3​
tan(x)=−2tan(60)3+9+8tan2(60)​​,tan(x)=2tan(60)8tan2(60)+9​−3​
tan(x)=−2tan(60)3+9+8tan2(60)​​:x=arctan(−2tan(60)3+9+8tan2(60)​​)+πn
tan(x)=−2tan(60)3+9+8tan2(60)​​
Apply trig inverse properties
tan(x)=−2tan(60)3+9+8tan2(60)​​
General solutions for tan(x)=−2tan(60)3+9+8tan2(60)​​tan(x)=−a⇒x=arctan(−a)+πnx=arctan(−2tan(60)3+9+8tan2(60)​​)+πn
x=arctan(−2tan(60)3+9+8tan2(60)​​)+πn
tan(x)=2tan(60)8tan2(60)+9​−3​:x=arctan(2tan(60)8tan2(60)+9​−3​)+πn
tan(x)=2tan(60)8tan2(60)+9​−3​
Apply trig inverse properties
tan(x)=2tan(60)8tan2(60)+9​−3​
General solutions for tan(x)=2tan(60)8tan2(60)+9​−3​tan(x)=a⇒x=arctan(a)+πnx=arctan(2tan(60)8tan2(60)+9​−3​)+πn
x=arctan(2tan(60)8tan2(60)+9​−3​)+πn
Combine all the solutionsx=arctan(−2tan(60)3+9+8tan2(60)​​)+πn,x=arctan(2tan(60)8tan2(60)+9​−3​)+πn
Show solutions in decimal formx=−1.46681…+πn,x=0.20575…+πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for 2tan(60-x)=tan(x) ?

    The general solution for 2tan(60-x)=tan(x) is x=-1.46681…+pin,x=0.20575…+pin
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