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Find all solutions in the interval [0, 2pi). give exact values and show all algebra. 2cos^2 X +3sinX-3=0

Mathematics
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trick is to rewrite \[cos^2(x)= 1-sin^2(x)\]
then get a quadratic equation in \[sin(x)\]
\[2cos^2(x)+3sin(x)-3=0\] \[2(1-sin^2(x))+3sin(x)-3=0\] \[2-2sin^2(x)+3sin(x)-3=0\] \[-2sin^2(x)+3sin(x)-1=0\] \[2sin^2(x)-3sin(x)-1=0\]

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Other answers:

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 \[z=1\]
then replace z by sin(x) and solve for x \[sin(x)=1\] \[x=\frac{\pi}{2}\]
\[sin(x)=\frac{1}{2}\] \[x=\frac{\pi}{6}\] or \[x=\frac{5\pi}{6}\]
ooops second answer is wrong sorry
\[sin(x)=\frac{1}{2}\] \[x=\frac{11\pi}{6}\]
k?
so x=11pi/6?
there are two places on the interval from 0 to 2pi where sine is 1/2
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
ohh i see thank you!
welcome!
any more?
huh?
any more of these to do?
well i just put some up
Does anyone happen to know 2cos^2 x+sinx-1=0 solve for x on [0,2pi)

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