If $\sqrt{3}a \cos x + 2b \sin x=c$ has 2 distinct real roots, then what is the range of $2c/(3a+2b)$? Let $a$,$b$,$c$ be three non-zero real numbers such that the equation $\sqrt{3}a\cos x + 2b \sin x = c$ ,($x$ belongs to $[-\pi/2,\pi/2]$) has two distinct real roots m and n with $m+n = \pi/3$. If range of values of $2c/(3a+2b)$ is $[q, r)$ then what is the value of $q+r$?
I tried to solve this question by putting $m$ and $n$ in the given equation and then subtracting the two equations. On solving I got the relation between $a$ and $b$ as $a=2b$.
On substituting the value of b in the given equation, and simplifying, I got
$\sin(x+\pi/3) = c/(2a)$
Also, $2c/(3a+2b) =c/(2a)$       (as $2b=a$)
For $x$ belongs to $[-\pi/2, \pi/2]$, $\sin(x+\pi/3)$ ranges from $-1/2$ to $1/2$. So, $q+r=0$.
But the answer is given as $1.5$. I'm unable to understand how? Someone please help me out with this question. Thank you

Well... think it reversely. If we write the two roots in advance, can we determine $a,b,c$? Of course, we can.
Let the roots to be $m,n$ so we have
$$\sqrt{3}a\cos m + 2b \sin m = c, \sqrt{3}a\cos n + 2b \sin n = c.$$
So we have $$\sqrt{3}a(\cos m-\cos n) + 2b (\sin m-\sin n)=0.$$
And using sum-to-difference we have $$\sqrt{3}a(\sin \frac{m-n}2\sin\frac{m+n}2) + 2b (\sin \frac{m-n}2\cos\frac{m+n}2)=0.$$
Since $m\ne n$, and $m+n=\frac\pi 3$ we have $$\sqrt{3}a(\frac 12) + 2b (\frac {\sqrt 3}2)=0.$$
or we have $a=-2b$.
So we have $3a+2b=2a$ and $$c/a=\sqrt{3}\cos m - \sin m.$$
And thus $$2c/(3a+2b)=2c/2a=c/a=\sqrt{3}\cos m - \sin m.$$
Notice that $m+n=\frac\pi 3$ and $m,n\le\frac\pi 2$, WLOG $m> n$, we have $\frac\pi6<m\le\frac\pi2$, and $\sqrt{3}\cos m - \sin m$ is clearly decreasing. So the range is $[-1,0)$. So $q+r=-1$
