1. ## Solve for x

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

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

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

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

Continue?.

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

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

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

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

Continue?.

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

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

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

4. ## Thanks....

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