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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.

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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.

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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 $-1$.

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ughhh here i am thinking its can be solved with the trig identities thanks – user66288 Mar 13 '13 at 17:03

$$ 1-2\sin^2A + \sin A = 2 $$ 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.

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"it's either of two numbers that are solutions of the quadratic equation". But $2x^2-x+1=0$ has $\Delta = 1 - 8 < 0$. – Jean-Claude Arbaut Mar 13 '13 at 17:12
Actually, Alpha shows there are no real solutions – Ross Millikan Mar 13 '13 at 17:15
Correct. So there are no solutions between $0$ and $2\pi$, nor anywhere on the real line. There are solutions if you want to talk about the sine of a complex number. – Michael Hardy Mar 13 '13 at 17:49

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