trig equation

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find x in radians

1. 2sin[squared]x+3cosx=0
 

happy

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Oct 30, 2004
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Well, what are your thoughts on this one since a similar one has been done for you? Where are you stuck?
 
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REPLY

I guess the different signs threw me off.
 

pka

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The problem can be rewritten as \(\displaystyle \begin{eqnarray*}
2\sin ^2 (x) + 3\cos (x) & = & 0 \\
2 - 2\cos ^2 (x) + 3\cos (x) & = & 0 \\
2\cos ^2 (x) - 3\cos (x) - 2 & = & 0 \\
\end{array}\).

Now can you solve \(\displaystyle w = \cos (x)\quad \Rightarrow \quad 2w^2 - 3w - 2 = 0\)
 
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(2w+1)(w-2)=0
2w+1=0 w-2=0

w=-1/2 w=2
 

tkhunny

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OK, now we have:

cos(x) = -1/2

1) Why did I throw out W = 2?
2) What are the solutions for x?
 
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tkhunny said:
OK, now we have:

cos(x) = -1/2

1) Why did I throw out W = 2?
2) What are the solutions for x?

because cos doesn't go to 2
 

tkhunny

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Excellent.

How about the other one?
 
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