What trig identity should i use for this

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0.15348=0.1415cosβ -0.291sinβcosβ

how do i solve this equation with both sinβ and cosβ, i realize that i need to play with the identities but have had no luck,
please help

i tried squaring the whole thing, and saying cos2β=t,
then i get

0.023556=0.02t - 0.582\sqrt{t-1}\sqrt{t} +0.08468t*(1-t)

then i have to square again to get rid of the rooy, but it seems to never end,
how else can i solve this to findβ
 
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One small thing you could do is to replace all of the fractional constants with variables, say A, B, C, and D. When you finally get solutions for t, substitute the numbers back in.

Looking at your equation, my sense is that you're going to get a 4th degree or higher equation in t, and there might not be a way to get exact solutions, but you can probably get approximate solutions to any desired precision.
 
can you not see any identities that will help me
 
None that spring readily to mind. I thought about this one in a different thread you had, but didn't get anywhere. One that comes to mind is sinb*cosb = 1/2 * sin 2b, but then you have the other cos b term, so that doesn't do any good.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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