Angle Help15 = arctan(2/x) - arctan (1/x)

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The discussion centers on solving the equation 15 = arctan(2/x) - arctan(1/x) for x. The initial approach involved using the tangent subtraction formula, leading to a quadratic equation that appeared to have no real roots. However, a correction revealed that the equation should have used tan(15) instead of 15, resulting in x = 0.64 and x = 3.08 as approximate solutions. Further analysis confirmed that rearranging the equation yields a quadratic with no real solutions. The final conclusion emphasizes the importance of correctly interpreting the problem to find valid solutions.
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Homework Statement


Basically, solve for x
15 = arctan(2/x) - arctan (1/x)

Homework Equations


tan (A-B) = (Tan A -Tan B) / (1+Tan A*Tan B)

The Attempt at a Solution


I really tried everything.
My first step was to:
Let y = arctan (2/x)
Therefore, tan y = 2/x
Similarly, u = 1/x
Then, tan (y-u) = (Tan y -Tan u) / (1+Tan u*Tan y) = 15
15 = (2/x-(1/x) / (1+(2/x^2)
15 = (1/x) / (x^2 + 2 / x^2)
15 = (x) / (x^2+2)
15x^2 + 30 - x = 0
Which has no real roots :(
But, with guess and check, it's around 0.65
 
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You need to post an attempt before we can help you. Start by taking the tangent of both sides.
 
Yeah, sorry. I just misclicked the first time
 
Alright, sorry guys to waste your time, but I believe I figured it out. Thanks for the hint of "tanning" both sides.
Instead of 15, it's supposed to be tan 15.
So that,
x/(x^2+2) = tan 15
x = 0.64
x = 3.08 (approximately)

Thanks for the help!
 
If you rearrange that equation you'll get the quadratic

x^2-\frac{1}{tan(15)}x+2=0

There are no real solutions to this quadratic.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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