trig equation

Well, what are your thoughts on this one since a similar one has been done for you? Where are you stuck?
 
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 \(\displaystyle w = \cos (x)\quad \Rightarrow \quad 2w^2 - 3w - 2 = 0\)
 
OK, now we have:

cos(x) = -1/2

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


because cos doesn't go to 2
 
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