# Trigonometric Equation: $4\sin\theta = 3\tan\theta$

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.

• 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 $A=0$. – David Mitra Apr 20 '19 at 9:41

Hint: Write $$4\sin(x)-3\tan(x)=0$$ and this is $$\sin(x)\left(4-\frac{3}{\cos(x)}\right)=0$$
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 $$-180^\circ \leq \theta \leq 180^\circ \implies m = 0$$.