# Math

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Find all solutions on the interval (0, 2pi) of the equation: 2(sin^2)t-3sint+1=0
how do you get this one started?

• Math -

let s=sin(t), so
2(sin^2)t-3sint+1=0
becomes
2s²-3s+1=0
(2s-1)(s-1)=0
=>
s=1/2 or s=1
=>
sin(x)=1/2 or sin(x)=1
Now solve for x for 0≤x≤2π

• Math -

That makes so much sense. Thank you!

• Math :) -

You're welcome!

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