anonymous
  • anonymous
Find all solutions in the interval [0, 2pi). give exact values and show all algebra. 2cos^2 X +3sinX-3=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
trick is to rewrite \[cos^2(x)= 1-sin^2(x)\]
anonymous
  • anonymous
then get a quadratic equation in \[sin(x)\]
anonymous
  • anonymous
\[2cos^2(x)+3sin(x)-3=0\] \[2(1-sin^2(x))+3sin(x)-3=0\] \[2-2sin^2(x)+3sin(x)-3=0\] \[-2sin^2(x)+3sin(x)-1=0\] \[2sin^2(x)-3sin(x)-1=0\]

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More answers

anonymous
  • anonymous
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
  • anonymous
then replace z by sin(x) and solve for x \[sin(x)=1\] \[x=\frac{\pi}{2}\]
anonymous
  • anonymous
\[sin(x)=\frac{1}{2}\] \[x=\frac{\pi}{6}\] or \[x=\frac{5\pi}{6}\]
anonymous
  • anonymous
ooops second answer is wrong sorry
anonymous
  • anonymous
\[sin(x)=\frac{1}{2}\] \[x=\frac{11\pi}{6}\]
anonymous
  • anonymous
k?
anonymous
  • anonymous
so x=11pi/6?
anonymous
  • anonymous
there are two places on the interval from 0 to 2pi where sine is 1/2
anonymous
  • anonymous
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
anonymous
  • anonymous
ohh i see thank you!
anonymous
  • anonymous
welcome!
anonymous
  • anonymous
any more?
anonymous
  • anonymous
huh?
anonymous
  • anonymous
any more of these to do?
anonymous
  • anonymous
well i just put some up
anonymous
  • anonymous
Does anyone happen to know 2cos^2 x+sinx-1=0 solve for x on [0,2pi)
anonymous
  • anonymous
@satellite73

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