Solving $5\sin x=1+2\cos^2x$, without sketching a graph $$5\sin x=1+2\cos^2x$$
"Solve, for $0 \le x <360$" I know using rearrangement $x= 30°$.
I also know using a graph sketch and knowledge of symmetry that the other solutions within the range can be derived; however, I'm under the impression that there's an algebraic way of concluding the solutions. I Recall my teacher somehow doing so but I dont remember how... Is the fastest way really to sketch out the $\sin$ graph and then visually work out the other solutions?
 A: I'm going to start with the fact that $\sin x=\frac 1 2$ since you told me you knew that in the comments. Once you have that, using a calculator, you should get $\sin^{-1} \frac 1 2=30^\circ$. Now, you know that $\sin(180^\circ-x)=\sin x$, so you also have the answer $180^\circ-30^\circ=150^\circ$. Again, no graphing required, but you still have to use an identity of $\sin(180^\circ-x)=\sin x$.
For $\cos x$, this identity is $\cos x=\cos -x=\cos(360^\circ-x)$. For $\tan x$, it is $\tan x=\tan(x+180^\circ)$. Using these identities, you can generate answers other than what your calculator gives you without graphing.
A: The well known identity: cos^2(x)=1-sin^2(x),
and the problem: 5sin(x)-1-2cos^2(x)=0,
lead to: 5sin(x)-1-2(1-sin^2(x)).
In other words (call sin(x) as Z): 5Z-1-2(1-Z^2)=0
This is a second degree equation. Apply the well known formula to solve.
Solve for Z, and the set x=arcsin(Z) to get the answerS.
    Recall that sin(x)->z means arcsin(z)->x. Arcus functions are the inverses.
Finally apply the degree limitation (0<=d<360) to choose the correct branch (closest to zero).
A: By the identity $\cos^2x+\sin^2x=1$, which holds for all $x$, your equation becomes
$$5\sin x=1+2(1-\sin^2x).$$
Rearranging the terms shows that this has the same solutions as
$$2\sin^2x+5\sin x-3=0.$$
This is a quadratic equation in $\sin x$, so the quadratic formula tells us that
$$\sin x=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot(-3)}}{2\cdot 2}=\frac{5\pm\sqrt{49}}{4},$$
so either $\sin x=\tfrac12$ or $\sin x=-3$. Of course the latter is impossible as $\sin x$ takes values from $-1$ to $1$, so $\sin x=\tfrac12$. In high school I was forced to learn by heart that then
$$x=30^{\circ}\qquad\text{ or }\qquad x=150^{\circ},$$
but I'm sure you have your own way of solving this.
