The problem is $2\cos t - 3\sin^2t +2 = 0$.
I get to $2\cos t -3\sin^2t =-2$
I think that I need to use a trigonometric identity like $\cos(x+y)$ and to divide $2\cos t -3\sin^2t$ with the $\sqrt{2^2+3^2}$
Do you know how to solve this? It should be $\sqrt{2^2 + 3^2}$
Answer
$$2 \cos t - 3 \sin^2t +2 = 0\\.$$ Write $$\sin^2 t=1-\cos^2 t, $$ then you have a quadratic equation. Solve for $\cos t$ $$ 2\cos t -3(1-\cos^2 t)+2=0\\3\cos ^2 t+2\cos t-3+2=0\\\cos t=-1,\;\frac{1}{3}. $$
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