TMUA practice: Trig Equations, problem 3

We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$

Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)$ , which can be derived using sine angle addition with $\sin{(2x + x)}$ , we have

$a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)$

Since $\sin{x} \ne 0$ as it is on the open interval $(0, \pi)$ , we can divide out $\sin{x}$ from both sides, leaving us with

$a\cdot(1+2\cos{x})=4\cos^2{x}-1$

Now, distributing $a$ and rearranging, we achieve the equation

$4\cos^2{x} - 2a\cos{x} - (1+a) = 0$

which is a quadratic in $\cos{x}$ .

Applying the quadratic formula to solve for $\cos{x}$ , we get

$\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}$

and expanding the terms under the radical, we get

$\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}$

Factoring, since $4a^2+16a+16 = (2a+4)^2$ , we can simplify our expression even further to

$\cos{x} =\frac{a\pm(a+2)}{4}$

Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$ .

Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \frac{2\pi}{3}$ , we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.

As $x\in (0, \pi)$ , $\cos{x}\in (-1, 1)$ , and therefore $\frac{a+1}{2}\in (-1, 1)$ , and $a\in(-3,1)$ .

There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$ , as this would lead to a double root for $\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \ne -2$ .

Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$ , so the answer is $-3-2+1 = \boxed{\textbf{(A) } {-}4}$ .

• DavidHovey