Answer:
third option: [tex]\frac{\pi }{2} , \frac{3\pi }{2} , \frac{\pi }{6} , and \frac{5\pi }{6}[/tex]
Step-by-step explanation:
We write the sin(2x) using the property of sin of the double angle:
[tex]sin(2\theta)=2*sin(\theta)*cos(\theta)[/tex] and replace the expression on the left of the given equation by this:
[tex]sin(2\theta)=cos(\theta)\\2*sin(\theta)*cos(\theta)=cos(\theta)[/tex]
now we notice that if [tex]cos(\theta)[/tex] equals zero, the equation becomes true. Therefore all of the values [tex]\theta[/tex] that make [tex]cos(\theta)-0[/tex] are solutions. That is [tex]\theta=\frac{\pi }{2} and \theta=\frac{3\pi }{2}[/tex].
Now, in the case [tex]cos(\theta)[/tex] is NOT zero, we can divide both sides of the equation by it, ending up with a simple answer:
[tex]2*sin(\theta)*cos(\theta)=cos(\theta)\\2*sin(\theta)=\frac{cos(\theta)}{cos(\theta)} \\2*sin(\theta)=1\\sin(\theta)=\frac{1}{2}[/tex]
and in the interval between 0 and [tex]2\pi[/tex] the solutions to this are:
[tex]\frac{\pi }{6} , and \frac{5\pi }{6}[/tex]
So we found a total of four solutions: [tex]\frac{\pi }{2} , \frac{3\pi }{2} , \frac{\pi }{6} , and \frac{5\pi }{6}[/tex]
