Question

Solve the following for the interval [0,2pi]:
6tan^2x+tanx=sqrt(3)tanx

Answer 1

please try this substitution:
tan x= z

Answer 2

\[6z^2+z =\sqrt{3}z\]\[6z^2+z -\sqrt{3}z=0\]\[6z^2 +(1-\sqrt{3})z =0\]\[z(6z +(1-\sqrt{3})) =0\]Since the product of two terms here is zero, one of the terms must be zero
\[z = 0\]or \[6z+1-\sqrt3= 0\]this would give us \[tan x = 0\] or \[6tan x= \sqrt3-1\]\[tan x=\frac{ \sqrt3-1}{6}\] since the interval is \[[0, 2\pi]\] thus we can now find the values of x in this interval for which \[tanx= 0\] and \[tanx=\frac{\sqrt 3-1}{6}\]