$$\tt{tan x =tan(x+10^0)tan(x+20^0)tan(x+30^0)}$$ This equation asked from a genius student of my class . I think and solve it like below . My idea based on $cot(x+30) tan(x)=tan(x+10)tan(x+20)\\15=\dfrac{0+30}{2}=\dfrac{10+20}{2}\\so $
$u=x+15^0$ and substitute $u$ in equation . Then my genius student asked for a other Idea to work with this .I show my work , but I can't go on ...
please help me to find an Idea . thanks in advanced .
$u=x+15 \to $ $$tan(u+15)=tan(u-5)tan(u+5)tan(u+15) \\ cot(u-15)tan(u+15)=tan(u-5)tan(u+5) \\ \dfrac{sin(u-15)cos(u+15)}{cos(u-15)sin(u+15)}= \dfrac{sin(u-5)sin(u+5)}{cos(u-5)cos(u+5)}$$ then by product to sum formulas we have $$\dfrac{-sin30 +sin 2u}{sin30 +sin 2u}=\dfrac{cos 10 -cos 2u}{cos10 +cos 2u}\\ \dfrac{2sin2u-1}{2sin2u+1}=\dfrac{cos 10 -cos 2u}{cos10 +cos 2u}\\1+\dfrac{2sin2u-1}{2sin2u+1}=1+\dfrac{cos 10 -cos 2u}{cos10 +cos 2u}\\\dfrac{4sin2u}{2sin2u+1}=\dfrac{2cos 10 }{cos10 +cos 2u}$$multiply and simplify $$ sin4u=cos 10\\ sin 4u=sin 80 \to \begin{cases}4u=80+2k\pi\\4u=180-80+2k\pi\end{cases}$$ then find $x=5^0+k\dfrac{\pi}{2} ,x=10^0+k\dfrac{\pi}{2}$