Question

Find all solutions in the interval [0, 2π).
7 tan3x - 21 tan x = 0
(I have no idea what to do)

Answer 1

tan(3x) can be expanded. Are you familiar with that trig identity?

Answer 2

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

Answer 3

\[
7\tan(3x) - 21\tan(x) = 0 \\
\text{Divide by 7:}\\
\tan(3x) - 3\tan(x) = 0 \\
\tan(3x) = 3\tan(x) \\
\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)} = 3\tan(x) \\
\text{Cross multiply, simplify and solve for x:} \\
3\tan(x)-\tan^3(x) = 3\tan(x)(1-3\tan^2(x)) \\
3\tan(x)-\tan^3(x) = 3\tan(x) - 9\tan^3(x)) \\
\text{Subtract 3 tan(x):} \\
-\tan^3(x) = - 9\tan^3(x)) \\
\text{Add 9 tan^3(x):}\\
8\tan^3(x) = 0 \\
\tan^3(x) = 0 \\
\tan(x) = 0 \\
x = 0, ~~\pi, ~~2\pi

\]