# Thread: [SOLVED] Number of solutions of arctan equation?

1. ## [SOLVED] Number of solutions of arctan equation?

Number of solutions of the equation arctan [1/(2x^2 + 1)] + arctan [1/(4x^2 + 1) = arctan 2/x^4 is

A)0
B)1
C)2
D)4

2. Hello, fardeen_gen!

Number of solutions of the equation: . $\arctan\left(\frac{1}{2x^2 + 1}\right) + \arctan\left(\frac{1}{4x^2 + 1}\right) \;= \;\arctan\left(\frac{2}{x^4}\right)$ is

. . $A)\;0 \qquad B)\;1 \qquad C)\;2 \qquad D)\;4$

Let: . $A = \arctan\left(\frac{1}{2x^2+1}\right),\; \;B = \arctan\left(\frac{1}{4x^2+1}\right),\;\;C = \arctan\left(\frac{2}{x^4}\right)$ .[1]

. . Hence: . $\tan A = \frac{1}{2x^2+1}\quad \tan B = \frac{1}{4x^2+1} \quad \tan C = \frac{2}{x^4}$ .[2]

Substitute [1] and we have: . $A + B \;=\;C$

Take the tangent of both sides:

. . $\tan (A + B) \;=\;\tan(C)$

. . $\frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)} \;=\;\tan(C)$

. . $\tan(A) + \tan(B) \;=\;\tan C - \tan(A)\tan(B)\tan(C)$

Substitute [2]:

. . $\frac{1}{2x^2+1} + \frac{1}{4x^2} \;=\;\frac{2}{x^4} - \frac{1}{2x^2+1}\cdot\frac{1}{4x^2+1}\cdot\frac{2} {x^4}$

Multiply by the LCD, $x^4(2x^2+1)(4x^2+1)$

. . $x^4(4x^2+1) + x^4(2x^2+1) \;=\;2(2x^2+1)(4x^2+1) - 2$

This simpifies to: . $6x^6 - 14x^4 - 12x^2 \:=\:0$

. . which factors: . $2x^2(3x^2 + 2)(x^2 - 3) \;=\;0$

Then: . $\begin{Bmatrix}
2x^2\:=\:0 & \Rightarrow & x \:=\:0 & \text{extraneous root} \\ \\[-4mm]
3x^2+2 \:=\:0 & \Rightarrow & x^2 \:=\:-\frac{2}{3} & \text{no real roots} \\ \\[-4mm]
x^2-3 \:=\:0 & \Rightarrow & x \:=\:\pm\sqrt{3} & \text{Yes!} \end{Bmatrix}$

There are two solutions . . . answer (C).