Rolle's Theorem $2+\sin(2x)-a\cdot\sin(x)\cos(x)=0$

Question

I have this equation: $2+\sin(2x)-a\cdot\sin(x)\cos(x)=0$

I need to find $a$ such that the equation has solutions.I tried with Rolle's Theorem:

I took $f(x)=2+\sin(2x)-a\cdot\sin(x)\cos(x)$ then $f'(x)=(2-a)\cdot\cos(2x)$ So if $f'(x)=0$ then $\cos(2x)=0$ so $x={+/-\frac{\pi}{4}+k\pi}$.

Then I made a table like in this picture. The right answer is $a\in(-\infty -2]\cup [6,\infty )$

Well, I have this answer.My sign is changing in that interval so I have a root, but what sign I have under $-\infty$?What about $\infty$?The limit as x approaches infinity doesn't exist, right?Then what sign should I have there?Because if I put $-$ under $-\infty$ I get my result is changing.

Maybe the domain is wrong, but I think it's $\mathbb R$, right? What's my mistake?

Answer 1

Just note that since $\sin x \cos x = \frac 12 \sin (2x)$, the equation becomes

$$ 2 + \sin (2x)-\frac a2 \sin (2x) = 0\Leftrightarrow \sin(2x) = \frac{-2}{1-\frac a2} = -\frac{4}{2-a} $$

You get solutions if $-1 \leq -\frac{4}{2-a} \leq 1$.