Answer:
[tex]\theta=k\pi, \hspace{3}k\in Z\\\\ or\\\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z\\[/tex]
Step-by-step explanation:
Factor constant terms:
[tex]-2(-4sin(2\theta)+sin(\theta))=0[/tex]
Divide both sides by -2:
[tex]sin(\theta)-4sin(2 \theta)=0[/tex]
Expand trigonometric functions using the fact:
[tex]sin(2 \theta) =2 sin(\theta) cos(\theta)[/tex]
So:
[tex]sin(\theta) -8sin(\theta)cos(\theta)=0[/tex]
Factor sin(x) and constant terms and multiply both sides by -1:
[tex]sin(\theta) (8cos(\theta)-1)=0[/tex]
Split into two equations:
[tex](1)=8cos(\theta)-1=0\\\\(2)=sin(\theta)=0[/tex]
For (1)
Add 1 to both sides and divide both sides by 8:
[tex]cos(\theta)=\frac{1}{8}[/tex]
Take the inverse cosine of both sides:
[tex]\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z[/tex]
For (2)
Simply take the inverse sine of both sides
[tex]\theta = k \pi, \hspace{3}k\in Z[/tex]
Therefore, the solutions are given by:
[tex]\theta=k\pi, \hspace{3}k\in Z\\\\ or\\ \theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z\\[/tex]
