1. ## Solve for x

Solve for x: sin x = cos (2x + 15)

2. $\displaystyle sin(x)=cos(2x+15)$

$\displaystyle cos^{-1}(sin(x))=2x+15$

$\displaystyle \frac{\pi}{2}-sin^{-1}(sin(x))=2x+15$

Continue?.

How many solutions are you looking for?. In a certain interval?. Just one?.

3. Originally Posted by galactus
$\displaystyle sin(x)=cos(2x+15)$

$\displaystyle cos^{-1}(sin(x))=2x+15$

$\displaystyle \frac{\pi}{2}-sin^{-1}(sin(x))=2x+15$

Continue?.

How many solutions are you looking for?. In a certain interval?. Just one?.
I'll just add a bit of background:

$\displaystyle \sin x = \cos \left(\frac{\pi}{2} - x \right)$ and so the equation can be written as $\displaystyle \cos \left(\frac{\pi}{2} - x \right) = \cos (2x + 15)$.

Now you need to understand the symmetry of the unit circle: $\displaystyle \cos A = \cos B$ means that either $\displaystyle A = B + 2 n \pi$ or $\displaystyle A = -B + 2 n \pi$ where n is an integer.

4. ## Thanks....

I want to thank both replies. I am learning a lot here.