Answer:
[tex]0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi[/tex]
Step-by-step explanation:
Given the equation:
[tex]sin(x)-2sin(x) tan(x)=0[/tex]
To find:
The solutions of given equation in the range [0, 2[tex]\pi[/tex]) i.e. 0 can be in the answer but [tex]2\pi[/tex] can not be there in the answer.
Solution:
Taking [tex]tan(x)[/tex] common, we get:
[tex]tan(x) (1-2sin(x))=0[/tex]
The solutions can be given as:
[tex]tan(x) = 0[/tex] OR [tex]1-2sin(x) = 0[/tex]
Let us solve both the equations one by one:
First equation:
[tex]tan(x) = 0\\ \Rightarrow x=0, \pi[/tex]
Second equation:
[tex]1-2sin(x) = 0\\\Rightarrow 2sinx=1\\\Rightarrow sinx=\dfrac{1}{2}\\\Rightarrow x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}[/tex]
Therefore, the answers as a comma separated list are:
[tex]0, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \pi[/tex]
