November 2nd, 2016,
As these are trigonometric problems, I have moved this thread accordingly.
Let's look at the first one:
$ \displaystyle 4\cos^2(\theta)+5\sin(\theta)=3$
Now, we should observe that if we use a Pythagorean identity:
$ \displaystyle \cos^2(\alpha)=1-\sin^2(\alpha)$
We will obtain an equation that is a quadratic in $\sin(\theta)$...
$ \displaystyle 4\left(1-\sin^2(\theta)\right)+5\sin(\theta)=3$
Arrange in standard form:
$ \displaystyle 4\sin^2(\theta)-5\sin(\theta)-1=0$
Now, what do you get when you apply the quadratic formula?