So they tell me solve $\cos2A+\sin A=2$.
Ok guys I found the problem :/ I was wondering why he gave us this question but its actually
$$\cos2A+\sin A=-2$$
and I simplified to $-2x^2+x+3=0$, which has solutions $x= 3/2$, $x=-1$
So I'm guessing then I go but this cannot be found so once again I am lost.
Since $\cos 2 A \leq 1$ and $\sin A \leq 1$, you must have equality in both. Equation in sine gives $A = {\pi \over 2} + 2 k \pi$, and then $\cos 2 A = \cos \pi = -1$, contradiction. There is thus no solution.
You have reached a quadratic equation in $\sin A$. Solve.
First rewrite your equation as $2\sin^2 A-\sin A +1=0$. Then, to make things look more familiar, let $x=\sin A$. So now you have the equation $2x^2-x+1=0$.
Of course there is no (real) solution.
And one can see this without computation! For the only way that we can have $\cos 2A+\sin A=2$ is if both $\sin A$ and $\cos 2A$ are equal to $1$.
The only place (in the interval from $0$ to $2\pi$, if you are using radians, or $0$ to $360$, if you are using degrees) is $\frac{\pi}{2}$ ($90$ degrees). But then the cosine of $2A$ is $-1$.
$$ 1-2\sin^2A + \sin A = 2 $$ Let $x=\sin A$. Then $$ 1-2x^2 + x = 2. $$ That is a quadratic equation. Once you've solved it for $x$, then you know what $\sin A$ is: it's either of two numbers that are solutions of the quadratic equation. Then there's the problem of finding $A$. If you just want solutions that you meet going around the circle once, you want solutions between $0$ and $2\pi$. You should find four of them: two for each of the two numbers that $\sin A$ could be.
