animalsavior94
Find all solutions in the interval [0, 2pi). give exact values and show all algebra.
2cos^2 X +3sinX-3=0
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anonymous
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trick is to rewrite
\[cos^2(x)= 1-sin^2(x)\]
anonymous
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then get a quadratic equation in \[sin(x)\]
anonymous
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\[2cos^2(x)+3sin(x)-3=0\]
\[2(1-sin^2(x))+3sin(x)-3=0\]
\[2-2sin^2(x)+3sin(x)-3=0\]
\[-2sin^2(x)+3sin(x)-1=0\]
\[2sin^2(x)-3sin(x)-1=0\]
anonymous
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just solve as a quadratic. if it is easier to see replace sin(x) by z to get
\[2z^2-3z+1=0\]
\[(2z-1)(z-1)=0\]
\[z=\frac{1}{2}\] or \[z=1\]
anonymous
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then replace z by sin(x) and solve for x
\[sin(x)=1\]
\[x=\frac{\pi}{2}\]
anonymous
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\[sin(x)=\frac{1}{2}\]
\[x=\frac{\pi}{6}\] or \[x=\frac{5\pi}{6}\]
anonymous
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ooops second answer is wrong sorry
anonymous
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\[sin(x)=\frac{1}{2}\]
\[x=\frac{11\pi}{6}\]
anonymous
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k?
animalsavior94
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so x=11pi/6?
anonymous
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there are two places on the interval from 0 to 2pi where sine is 1/2
anonymous
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one is at
\[\frac{\pi}{6}\] and the other is at
\[\frac{11\pi}{6}\] so all together you have 3 answers.
those two and also
\[\frac{\pi}{2}\] where sine is 1
animalsavior94
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ohh i see thank you!
anonymous
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welcome!
anonymous
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any more?
animalsavior94
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huh?
anonymous
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any more of these to do?
animalsavior94
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well i just put some up
Gabysolis49
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Does anyone happen to know 2cos^2 x+sinx-1=0 solve for x on [0,2pi)
Gabysolis49
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@satellite73