How do I find all the solutions for cos(x)tan(x)=tan(x) on one interval rotation of the unit circle?

1 Answer
Oct 19, 2017

In [0, 2pi)[0,2π), we have x in 0 and pix0andπ

Explanation:

We can immediately rewrite as

0 = tanx - cosxtanx0=tanxcosxtanx

0 = tanx(1 - cosx)0=tanx(1cosx)

So we now have two equations.

tanx = 0tanx=0

x = 0 or pix=0orπ

AND

1 - cosx= 01cosx=0

x = 0x=0

So in the interval [0, 2pi)[0,2π), the solutions will be x in 0 uu pix0π

Hopefully this helps!