G

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find x in radians

1. 2sin[squared]x+3cosx=0

1. 2sin[squared]x+3cosx=0

G

find x in radians

1. 2sin[squared]x+3cosx=0

1. 2sin[squared]x+3cosx=0

G

I guess the different signs threw me off.

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- Jan 29, 2005

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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 \(\displaystyle w = \cos (x)\quad \Rightarrow \quad 2w^2 - 3w - 2 = 0\)

G

(2w+1)(w-2)=0

2w+1=0 w-2=0

w=-1/2 w=2

2w+1=0 w-2=0

w=-1/2 w=2

G

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