Question

Solve sin 2ø= cos ø on the interval 0

Answer 1

Answer:

third option: $\frac{\pi }{2} , \frac{3\pi }{2} , \frac{\pi }{6} , and \frac{5\pi }{6}$

Step-by-step explanation:

We write the sin(2x) using the property of sin of the double angle:

$sin(2\theta)=2*sin(\theta)*cos(\theta)$ and replace the expression on the left of the given equation by this:

$sin(2\theta)=cos(\theta)\\2*sin(\theta)*cos(\theta)=cos(\theta)$

now we notice that if $cos(\theta)$ equals zero, the equation becomes true. Therefore all of the values $\theta$ that make $cos(\theta)-0$ are solutions. That is $\theta=\frac{\pi }{2} and \theta=\frac{3\pi }{2}$.

Now, in the case  $cos(\theta)$ is NOT zero, we can divide both sides of the equation by it, ending up with a simple answer:

$2*sin(\theta)*cos(\theta)=cos(\theta)\\2*sin(\theta)=\frac{cos(\theta)}{cos(\theta)} \\2*sin(\theta)=1\\sin(\theta)=\frac{1}{2}$

and in the interval between 0 and $2\pi$ the solutions to this are:

$\frac{\pi }{6} , and \frac{5\pi }{6}$

So we found a total of four solutions: $\frac{\pi }{2} , \frac{3\pi }{2} , \frac{\pi }{6} , and \frac{5\pi }{6}$