old problem question 118120 tan tan x =tan 3x −tan 2x let t=tan x (1) tan t =((t^5 +2t^3 +t)/(3t^4 −4t^2 +1)) for t≥0 we get (approximating) t_0 =0 t_1 ≈1.28941477 t_2 ≈4.17629616 t_3 ≈7.49316173 t_4 ≈10.7303610 t_5 ≈13.9285293 ... x=nπ+arctan t let n=0 to stay in the first period 0≤t<+∞ ⇒ 0≤arctan t <(π/2) ⇒ (1) has infinite solutions for 0≤x<(π/2) graphically this is easy to see, plot these: f_1 (t)=tan t f_2 (t)=((t(t^4 +2t^2 +1))/(3t^4 −4t^2 +1))=(t/3)+((2t(5t^2 +1))/(3(3t^4 −4t^2 +1))) ⇒ g(t)=(1/3)t is asymptote of f_1 (t) and obviously tan t =at with a∈R has infinite solutions


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Question Number 118371 by MJS_new last updated on 17/Oct/20

$old problem q u e s t i o n 118120$ $\tan \tan x = \tan 3 x - \tan 2 x$ $let t = \tan x$ $(1) \tan t = \frac{t^{5} + 2 t^{3} + t}{3 t^{4} - 4 t^{2} + 1}$ $for t ⩾ 0 we get (approximating)$ $t_{0} = 0$ $t_{1} \approx 1.28941477$ $t_{2} \approx 4.17629616$ $t_{3} \approx 7.49316173$ $t_{4} \approx 10.7303610$ $t_{5} \approx 13.9285293$ $. . .$ $x = n π + \arctan t$ $let n = 0 to stay in the first period$ $0 ⩽ t < + \infty \Rightarrow 0 ⩽ \arctan t < \frac{π}{2}$ $\Rightarrow (1) has infinite solutions for 0 ⩽ x < \frac{π}{2}$ $graphically this is easy to see, plot these :$ $f_{1} (t) = \tan t$ $f_{2} (t) = \frac{t (t^{4} + 2 t^{2} + 1)}{3 t^{4} - 4 t^{2} + 1} = \frac{t}{3} + \frac{2 t (5 t^{2} + 1)}{3 (3 t^{4} - 4 t^{2} + 1)}$ $\Rightarrow g (t) = \frac{1}{3} t is asymptote of f_{1} (t)$ $and obviously \tan t = a t with a \in R has$ $infinite solutions$
Commented by prakash jain last updated on 17/Oct/20

$Agree . There were mistakes in$ $my previous calculation .$
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