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Popular Trigonometry >

arctan(x)+arctan(2x)= pi/4

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Solution

arctan(x)+arctan(2x)=4π​

Solution

x=417​−3​
Solution steps
arctan(x)+arctan(2x)=4π​
Rewrite using trig identities
arctan(x)+arctan(2x)
Use the Sum to Product identity: arctan(s)+arctan(t)=arctan(1−sts+t​)=arctan(1−x⋅2xx+2x​)
arctan(1−x⋅2xx+2x​)=4π​
Apply trig inverse properties
arctan(1−x⋅2xx+2x​)=4π​
arctan(x)=a⇒x=tan(a)1−x⋅2xx+2x​=tan(4π​)
tan(4π​)=1
tan(4π​)
Use the following trivial identity:tan(4π​)=1
tan(4π​)
tan(x) periodicity table with πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​tan(x)033​​13​±∞−3​−1−33​​​​
=1
=1
1−x⋅2xx+2x​=1
1−x⋅2xx+2x​=1
Solve 1−x⋅2xx+2x​=1:x=−43+17​​,x=417​−3​
1−x⋅2xx+2x​=1
Simplify 1−x⋅2xx+2x​:1−2x23x​
1−x⋅2xx+2x​
Add similar elements: x+2x=3x=1−2xx3x​
1−x⋅2x=1−2x2
1−x⋅2x
x⋅2x=2x2
x⋅2x
Apply exponent rule: ab⋅ac=ab+cxx=x1+1=2x1+1
Add the numbers: 1+1=2=2x2
=1−2x2
=1−2x23x​
1−2x23x​=1
Multiply both sides by 1−2x2
1−2x23x​=1
Multiply both sides by 1−2x21−2x23x​(1−2x2)=1⋅(1−2x2)
Simplify
1−2x23x​(1−2x2)=1⋅(1−2x2)
Simplify 1−2x23x​(1−2x2):3x
1−2x23x​(1−2x2)
Multiply fractions: a⋅cb​=ca⋅b​=1−2x23x(1−2x2)​
Cancel the common factor: 1−2x2=3x
Simplify 1⋅(1−2x2):1−2x2
1⋅(1−2x2)
Multiply: 1⋅(1−2x2)=(1−2x2)=(1−2x2)
Remove parentheses: (a)=a=1−2x2
3x=1−2x2
3x=1−2x2
3x=1−2x2
Solve 3x=1−2x2:x=−43+17​​,x=417​−3​
3x=1−2x2
Switch sides1−2x2=3x
Move 3xto the left side
1−2x2=3x
Subtract 3x from both sides1−2x2−3x=3x−3x
Simplify1−2x2−3x=0
1−2x2−3x=0
Write in the standard form ax2+bx+c=0−2x2−3x+1=0
Solve with the quadratic formula
−2x2−3x+1=0
Quadratic Equation Formula:
For a=−2,b=−3,c=1x1,2​=2(−2)−(−3)±(−3)2−4(−2)⋅1​​
x1,2​=2(−2)−(−3)±(−3)2−4(−2)⋅1​​
(−3)2−4(−2)⋅1​=17​
(−3)2−4(−2)⋅1​
Apply rule −(−a)=a=(−3)2+4⋅2⋅1​
Apply exponent rule: (−a)n=an,if n is even(−3)2=32=32+4⋅2⋅1​
Multiply the numbers: 4⋅2⋅1=8=32+8​
32=9=9+8​
Add the numbers: 9+8=17=17​
x1,2​=2(−2)−(−3)±17​​
Separate the solutionsx1​=2(−2)−(−3)+17​​,x2​=2(−2)−(−3)−17​​
x=2(−2)−(−3)+17​​:−43+17​​
2(−2)−(−3)+17​​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅23+17​​
Multiply the numbers: 2⋅2=4=−43+17​​
Apply the fraction rule: −ba​=−ba​=−43+17​​
x=2(−2)−(−3)−17​​:417​−3​
2(−2)−(−3)−17​​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅23−17​​
Multiply the numbers: 2⋅2=4=−43−17​​
Apply the fraction rule: −b−a​=ba​3−17​=−(17​−3)=417​−3​
The solutions to the quadratic equation are:x=−43+17​​,x=417​−3​
x=−43+17​​,x=417​−3​
Verify Solutions
Find undefined (singularity) points:x=2​1​,x=−2​1​
Take the denominator(s) of 1−x⋅2xx+2x​ and compare to zero
Solve 1−x⋅2x=0:x=2​1​,x=−2​1​
1−x⋅2x=0
Move 1to the right side
1−x⋅2x=0
Subtract 1 from both sides1−x⋅2x−1=0−1
Simplify−x⋅2x=−1
−x⋅2x=−1
Simplify−2x2=−1
Divide both sides by −2−2−2x2​=−2−1​
x2=21​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
x=21​​,x=−21​​
21​​=2​1​
21​​
Apply radical rule: ba​​=b​a​​,a≥0,b≥0=2​1​​
Apply radical rule: 1​=11​=1=2​1​
−21​​=−2​1​
−21​​
Apply radical rule: ba​​=b​a​​,a≥0,b≥0=−2​1​​
Apply radical rule: 1​=11​=1=−2​1​
x=2​1​,x=−2​1​
The following points are undefinedx=2​1​,x=−2​1​
Combine undefined points with solutions:
x=−43+17​​,x=417​−3​
x=−43+17​​,x=417​−3​
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into arctan(x)+arctan(2x)=4π​
Remove the ones that don't agree with the equation.
Check the solution −43+17​​:False
−43+17​​
Plug in n=1−43+17​​
For arctan(x)+arctan(2x)=4π​plug inx=−43+17​​arctan(−43+17​​)+arctan(2(−43+17​​))=4π​
Refine−2.35619…=0.78539…
⇒False
Check the solution 417​−3​:True
417​−3​
Plug in n=1417​−3​
For arctan(x)+arctan(2x)=4π​plug inx=417​−3​arctan(417​−3​)+arctan(2⋅417​−3​)=4π​
Refine0.78539…=0.78539…
⇒True
x=417​−3​

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Frequently Asked Questions (FAQ)

  • What is the general solution for arctan(x)+arctan(2x)= pi/4 ?

    The general solution for arctan(x)+arctan(2x)= pi/4 is x=(sqrt(17)-3)/4
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