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3sint-2=cos^2 t
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cos^2t = 1-sin^2t Substitute and solve.
You get: sin^2 - 3sint + 1 =0
relpace cos^2 t by 1 - sin^2 t and solve the quadratic
I get: \[sin^2t+3sint-3=0\], not \[sin^2-3sint+1=0\]
\[3\sin t=2 \cos^2 t\]
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Yeah, right. Whatever. Re-arrange and solve is the point.
3sint-2=cos^2 t 3sint - 2 = 1 - sin^2 t sin^2 t + 3 sin t - 3 = 0
you'll need to use the quadratic formula to solve this or complete square if you prefer
ok thx
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