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How would you find all the solutions to this question:

Question

Solve this equation for -180° ≤ θ ≤ 180°. Show your working.

$4\sin\theta = 3\tan\theta$


My Solution

$$4\sin\theta = 3\tan\theta\\ \sin\theta = 3\frac{\sin\theta}{\cos\theta}\\ 4\sin\theta\cos\theta = 3\sin\theta\\ 4\cos\theta = 3\\ \cos\theta = \frac{3}{4}\\ \theta = 41.1°\ (to\ 1\ decimal\ place)$$

I know from the graphs of sine and tangent that 0°, 180°, -180° are also solutions to this equation but how do I show that these three are also solutions without the graphs (that is, in a similar way to how I showed that 41.1° is one solution)?

Thanks.

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  • 2
    $\begingroup$ Those are the solutions that correspond to $\sin\theta=0$. In the first step of your solution, you lost these upon the division. Moral: if you divide both sides of an equation by $A$, check what goes on when $A=0$. $\endgroup$ – David Mitra Apr 20 at 9:41
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Hint: Write $$4\sin(x)-3\tan(x)=0$$ and this is $$\sin(x)\left(4-\frac{3}{\cos(x)}\right)=0$$

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As David Mitra observed in the comments, by dividing through by $\sin\theta$, you lost those solutions in which $\sin\theta = 0$. You also failed to find another solution by failing to consider symmetry. \begin{align*} 4\sin\theta & = 3\tan\theta\\ 4\sin\theta - 3\tan\theta & = 0\\ 4\sin\theta - 3\frac{\sin\theta}{\cos\theta} & = 0\\ 4\sin\theta\cos\theta - 3\sin\theta & = 0\\ \sin\theta(4\cos\theta - 3) & = 0 \end{align*} \begin{align*} \sin\theta & = 0 & 4\cos\theta - 3 & = 0\\ \theta & = n\pi, n \in \mathbb{Z} & 4\cos\theta & = 3\\ & & \cos\theta & = \frac{3}{4}\\ & & \theta & = \pm \arccos\left(\frac{3}{4}\right) + 2m\pi, m \in \mathbb{Z} \end{align*} The requirement that $-180^\circ \leq \theta \leq 180^\circ$ is equivalent to $-\pi \leq \theta \leq \pi$. Hence, we must take $n = -1, 0, 1$, which yields $\theta = -\pi, 0, \pi$ or, equivalently, $\theta = -180^\circ, 0^\circ, 180^\circ$.

Since the function $f: [-1, 1] \to [0, \pi]$ defined by $f(x) = \arccos x$ yields the unique angle $\theta$ in the interval $[0, \pi]$ such that $\cos\theta = x$, there is one angle in the interval $[0, \pi]$ such that $\theta = \arccos\left(\frac{3}{4}\right)$. It is $\approx 0.7227342478$ radians, which converts to $\approx 41.4^\circ$ according to this calculator. However, we want all the solutions in the interval $-180^\circ \leq \theta \leq 180^\circ$. Since the cosine function is even, $\cos(-x) = \cos x$. Thus, $\theta = -\arccos\left(\frac{3}{4}\right) \approx -41.4^\circ$ is also a solution. There are no other solutions in the interval since $-180^\circ \leq \theta \leq 180^\circ \implies m = 0$.

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