Can You Solve this Trig Inverse Equation with arctan and arctan?

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To solve the equation arctan(3x) + arctan(2x) = π/4, the identity arctan[x] + arctan[y] = arctan[(x+y)/(1-xy)] is applied. This simplifies to finding the angle whose tangent is 1, leading to the equation 5x/(1-6x^2) = 1. Solving this results in x = 1/6 and x = -1, but -1 is rejected since it falls outside the domain of arctan. Therefore, the valid solution is x = 1/6.
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Homework Statement


Solve arctan 3x + arctan 2x= pie/4?


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The Attempt at a Solution

 
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there is a trig identiy that helps.

arctan[x]+arctan[y] = arctan[(x+y)/(1-xy)]

then it is just the angle that has a tan of pi/4 (i am assuming radians), so 1

then solving 5x/(1-6x^2)=1 we get x= 1/6 and x=-1
reject the -1 as the domain of arctan does not include it if is re-inserted into the fuction.
 

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