Posted by Kenz on Wednesday, May 30, 2012 at 10:13pm.
Find all solutions on the interval (0, 2pi) of the equation: 2(sin^2)t3sint+1=0
how do you get this one started?

Math  MathMate, Wednesday, May 30, 2012 at 10:38pm
let s=sin(t), so
2(sin^2)t3sint+1=0
becomes
2s²3s+1=0
(2s1)(s1)=0
=>
s=1/2 or s=1
=>
sin(x)=1/2 or sin(x)=1
Now solve for x for 0≤x≤2π

Math  Kenz, Thursday, May 31, 2012 at 9:52pm
That makes so much sense. Thank you!

Math :)  MathMate, Friday, June 1, 2012 at 7:44am
You're welcome!
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