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 $2/3$ ?

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
