Solve for x

**magentarita** · December 31st 2008, 02:31 PM

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

**galactus** · December 31st 2008, 02:59 PM

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

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

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

Continue?.

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

**mr fantastic** · December 31st 2008, 03:21 PM

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.

**magentarita** · January 1st 2009, 12:33 PM

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

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