Can we solve these trig equations for $x, y$ and $z$?

Can we solve these trig equations for $x, y$ and $z$?

Question

Find all the angles between $0^\circ$ and $180^\circ$ which satisfy the equations

1. $\tan (x + 60^\circ) = 1$,

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1. $8 \sin y + 3 \sec y = 0$,

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1. $3 \sin^2 z - 8 \sin z \cos z - 3 \cos^2 z = 0$.

Solution

Let’s find the general solution. We know that $\arctan{1} = 45 ^\circ$, so $x+60=45+180n \implies x = -15 + 180n.$

Recall that the function $\tan x$ is periodic with period $180^\circ$.

Thus the only solution in our range is $165^\circ$.

By definition,

$\sec$

is the reciprocal of

$\cos$

. By multiplying through by

$\cos y$

and rearranging, we have

$$\begin{equation} 8 \cos y \sin y = -3. \end{equation}$$

Note $\cos y \neq 0$, since then $\sec y$ would be undefined.

Now $2\sin y\cos y = \sin(2y)$, and so on rearranging, our equation becomes $\sin(2y) = -\frac{3}{4}.$

So the general solution here is $2y = -48.6 + 360n$, or $228.6+360n$.

This is since $228.6 = 180 - (-48.6)$.

This implies $y = -24.3 + 180n$, or $114.3 + 180n$, and so the solutions in our range are $114.3^\circ$ and $155.7^\circ$.

Could $\cos z$ be $0$ here? This would mean that for the equation to hold, $\sin z = 0$. But $\cos z$ and $\sin z$ are never zero together, so $\cos z$ cannot be zero if the equation holds.

So we can divide the equation by $\cos^2 z$, giving us $3 \tan^2 z - 8 \tan z - 3 = 0$, which becomes $(3\tan z + 1)(\tan z -3) = 0$.

So $\tan z = -\dfrac{1}{3}$ or $\tan z = 3$, and $z = -18.4 + 180n$ or $71.6 + 180n$.

Thus the two solutions in our range are $71.6^\circ$ and $161.6^\circ$.

Alternatively, we could factorise the original expression, $3 \sin^2 z - 8 \sin z \cos z - 3 \cos^2 z =(3\sin z+\cos z)(\sin z-3\cos z)$ with the same result.