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 $2/3$ ?

you have $$-3\sin(t)^2+5\sin(t)+2=0$$ let $$\sin(t)=u$$ then we have $$-3u^2+5u+2=0$$ divided by $-3$ gives $$u^2-\frac{5}{3}u-\frac{2}{3}=0$$
solving this equation we obtain
$$u_{1,2}=\frac{5}{6}\pm\sqrt{\frac{25}{36}+\frac{24}{36}}$$
from here you will come to the result.
