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anonymous
 5 years ago
Find all solutions in the interval [0, 2pi). give exact values and show all algebra.
2cos^2 X +3sinX3=0
anonymous
 5 years ago
Find all solutions in the interval [0, 2pi). give exact values and show all algebra. 2cos^2 X +3sinX3=0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0trick is to rewrite \[cos^2(x)= 1sin^2(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then get a quadratic equation in \[sin(x)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2cos^2(x)+3sin(x)3=0\] \[2(1sin^2(x))+3sin(x)3=0\] \[22sin^2(x)+3sin(x)3=0\] \[2sin^2(x)+3sin(x)1=0\] \[2sin^2(x)3sin(x)1=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just solve as a quadratic. if it is easier to see replace sin(x) by z to get \[2z^23z+1=0\] \[(2z1)(z1)=0\] \[z=\frac{1}{2}\] or \[z=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then replace z by sin(x) and solve for x \[sin(x)=1\] \[x=\frac{\pi}{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[sin(x)=\frac{1}{2}\] \[x=\frac{\pi}{6}\] or \[x=\frac{5\pi}{6}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ooops second answer is wrong sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[sin(x)=\frac{1}{2}\] \[x=\frac{11\pi}{6}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are two places on the interval from 0 to 2pi where sine is 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one 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
 5 years ago
Best ResponseYou've already chosen the best response.0any more of these to do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i just put some up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Does anyone happen to know 2cos^2 x+sinx1=0 solve for x on [0,2pi)
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