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 $2/3$ ?
 A: $$3\cos ^2t + 5\sin t = 1$$
$$3(1-\sin ^2t) + 5\sin t -1 = 0$$
$$-3\sin ^2t + 5\sin t + 2 = 0$$
$$3\sin ^2t - 5\sin t - 2 = 0$$
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?
A: 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.
A: We have, $$3\cos ^2t + 5\sin t = 1$$ $$\implies 3(1-\sin ^2t) + 5\sin t = 1$$ $$\implies 3\sin ^2t+ 5\sin t-2=0$$ Factorizing the expression, we get $$ (3\sin t-1)(\sin t+2)=0$$ $$\text{if}\ 3\sin t-1=0 \implies \sin t=\frac{1}{3}$$$$\implies \color{blue}{t=2n\pi+\sin^{-1}\left(\frac{1}{3}\right)}$$ $$\text{Or} \ \color{blue}{t=(2n+1)\pi-\sin^{-1}\left(\frac{1}{3}\right)}$$ $$\text{if}\ \sin t+2=0 \implies \sin t\neq -2$$ Hence, the general solution is $t=2n\pi+\sin^{-1}\left(\frac{1}{3}\right)$ or $t=(2n+1)\pi-\sin^{-1}\left(\frac{1}{3}\right)$ 
Where, n is any natural number
