Question

root 3 tan 3 theta -1 = 0

Answer 1

\[\sqrt[3]{\tan^3(\theta)}-1=0\]

Answer 2

oop I chould have been more clear Im \[\sqrt{3}\tan3\theta-1=0\] looking for all solutions in the interval [0,2pi)

Answer 3

isolate the tan(3theta) part

Answer 4

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

Answer 5

So let's replace u with 3 theta
that is set u=3 theta
we want to put our equation and our inequality in terms of the new variable

\[\tan(3 \theta)=\frac{1}{\sqrt{3} } , 0 \le \theta <2 \pi \\ \tan(u)=\frac{1}{\sqrt{3}}, 0 \le \frac{u}{3} <2 \pi \text{ since } \theta=\frac{u}{3} \\ \text{ now we want to solve the following equation satisfying the following inequality } \\ \tan(u)=\frac{1}{\sqrt{3}}, 0 \le u <6 \pi \text{ here I multiplied both sides of the inequality by } \\ 3 \text{ \to solve it for } u\]

Answer 6

@cahdmus You have become quiet. Have you finished this or just speechless?

Answer 7

I got \[(7\pi/18)-(k/3)\pi\] im jus still unsure how to do the interval