Trigonometric equation with sine and cosine So the equation is $3\cos ^2t + 5\sin t = 1$
Now I have simplified this to $$3(1-\sin ^2t) + 5\sin t -1 = 0$$
which leads to $$-3\sin ^2t + 5\sin t + 2 = 0$$
Then I get $$-3t^2 + 5 t +2 = 0$$
Is this the correct way to go with this equation then use $t = t/2 \pm \sqrt {(t/2)^ 2 + y}$ where $y$ in this case will be $2/3$ ?

$$3\cos ^2t + 5\sin t = 1$$
$$3(1-\sin ^2t) + 5\sin t -1 = 0$$
$$-3\sin ^2t + 5\sin t + 2 = 0$$
$$3\sin ^2t - 5\sin t - 2 = 0$$
Now , Let $\sin t =x$, then equation reduces to,
$$3x^2-5x-2=0$$
$$(3x+1)(x-2)=0$$, Now can you finish from here?
