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How do you solve #tan^-1(2x)+tan^-1(x)= (3pi)/17#?

Answer 1

#x = pi/17#

#tan^-1 (2x) #--> arc (2x) #tan ^-1 x # --> arc x The equation is transformed to: #2x + x = 3x = (3pi)/17# #x = pi/17# Answers in degrees: #pi/17 = 10^@59# #(2pi)/17 = 21^@18#

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Answer 2

Charles Autterson

Let #c = tan ((3pi)/17) = 0.619174#, nearly. Then, #x =(-3+- sqrt(9+8c^2))/(4c)=0.19129 and -2.61367#, nearly It is seemingly Incredible that x could be negative.

Let #a = tan^(-1)(2x) and b = tan^(-1) (x)#.

Then #tan (a + b ) = (tan a + tan b )/(1-tna tan b )#

#=(2x+x)/(1-(2x)(x))#

#=(3x)/(1-2x^2) = tan ((3pi)/17) = c=0.619174#, nearly.

Solving this quadratic for a positive x,

#x = (sqrt(9+8c^2)-3)/(4c)=0.19129#, nearly.

Here, the two angles that add up to #(3pi)/17 =31.76^o# are #20.94^o

and 10.83^o#.

Seemingly incredible but true, the negative root #=-2.61367# is

also a solution.

The corresponding angles are #100.83^o and -69.06^o#.

I have used that tangent is negative in the 2nd quadrant for

#tan^(-1)(2x)# and in the fourth quadrant for #tan^(-1)(x)#, respectively.

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Answer 3

Abigail Armstrong

To solve the equation ( \tan^{-1}(2x) + \tan^{-1}(x) = \frac{3\pi}{17} ), we can use the identity for the addition of inverse tangents:

[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) ]

Applying this identity to our equation, we have:

[ \tan^{-1}(2x) + \tan^{-1}(x) = \tan^{-1}\left(\frac{2x + x}{1 - 2x \cdot x}\right) ]

[ \tan^{-1}(2x) + \tan^{-1}(x) = \tan^{-1}\left(\frac{3x}{1 - 2x^2}\right) ]

So, our equation becomes:

[ \tan^{-1}\left(\frac{3x}{1 - 2x^2}\right) = \frac{3\pi}{17} ]

To solve for ( x ), we take the tangent of both sides:

[ \tan\left[\tan^{-1}\left(\frac{3x}{1 - 2x^2}\right)\right] = \tan\left(\frac{3\pi}{17}\right) ]

[ \frac{3x}{1 - 2x^2} = \tan\left(\frac{3\pi}{17}\right) ]

Now, we solve for ( x ):

[ 3x = (1 - 2x^2) \tan\left(\frac{3\pi}{17}\right) ]

[ 3x = \tan\left(\frac{3\pi}{17}\right) - 2x^2 \tan\left(\frac{3\pi}{17}\right) ]

[ 2x^2 \tan\left(\frac{3\pi}{17}\right) + 3x - \tan\left(\frac{3\pi}{17}\right) = 0 ]

This is now a quadratic equation in ( x ), which can be solved using the quadratic formula. Once we find the solutions for ( x ), we can check them to ensure they are valid solutions for the original equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.