Question

Find the solutions in the interval [0, 2π).

3 tan^3 θ − 3 tan^2 θ − tan θ + 1 = 0

Answer 1

one is
tan thita =1

(by hit and trial method)

Answer 2

I don't understand.

Answer 3

just put tan thita = 1 and u will realise that it is indeed a solution to the eqn

Answer 4

then to find the other 2 solutions just divide the equation with tan thita -1
and get a quadratic in tan thita,

which will give the other two roots

Answer 5

\[3u^3-3u^2-u+1=0\]
\[3u^2(u-1)-1(u-1)=0\]
\[(u-1)(3u^2-1)=0\]
\[=> u=1 \text{ or } u=\pm \sqrt{\frac{1}{3}}\]

Answer 6

\[\tan(\theta)=1 \text{ or } \tan(\theta)=\pm \frac{1}{\sqrt{3}}\]