# Solve, for $0 \leq x < 360$, $5\sin 2x = 2\cos 2x$. What about $\cos \theta = 0$?

Solve $5\sin 2x = 2\cos 2x$ for $0 \leq x < 360^\circ$.

Let $\theta = 2x$. Then $(5\tan \theta -2)\cos \theta = 0.$

So $\tan \theta = \dfrac 2 5$ or $\cos \theta = 0.$

Calculating the above value gives:

$\theta = 45.0^\circ, 135.0^\circ, 225.0^\circ, 315.0^\circ, 10.9^\circ, 100.9^\circ, 190.9^\circ$, or $280.9^\circ.$

But looking at the marking scheme strictly states:

$x = 10.9, 100.9, 190.9, 280.9$ (Allow awrt)
Extra solution(s) in range: Loses the final A mark.

As you can see this is only solution to the $\tan \theta = \dfrac 5 2$.
What am I missing here? Why is $\cos \theta = 0$ not a correct solution?

Is factorising $\cos \theta$ unnecessary because when $\cos \theta = 0$, $\tan \theta = \dfrac {\sin \theta}{\cos \theta}$ is undefined so that renders the equation useless?

• you have extra solutions. if $\cos x = 0,$ then $\sin x = 2 \cos x = 0.$ but then these $\sin x = 0 , \cos x = 0$ have no solution because $\sin^2 x + \cos ^2 x = 1.$ – abel May 18 '15 at 6:18
• When you "took out" $\cos\theta$, you were essentially dividing by $\cos\theta$. This is illegitimate if $\cos\theta=0$, and in this case introduces extraneous solutions. – André Nicolas May 18 '15 at 6:27
• Note you can do this without dividing by writing $c=\cos \theta, s=\sin \theta$ and $10cs=2c^2-2s^2$ or $4c^2-20cs-4s^2=(2c-5s)^2-29s^2=0$ so that $2c-5s=\pm\sqrt {29}s$ etc. You can't have $c=s=0$ as observed elsewhere. – Mark Bennet May 18 '15 at 6:51
Another method if you are interested. $5 \sin(2x)-2 \cos(2x)=0 \\ \frac{5}{\sqrt{29}} \sin(2x)-\frac{2}{\sqrt{29}} \cos(2x)=0 \\ \sin(\theta)\sin(2x)-\cos(\theta)\cos(2x)=0 \\ \cos(\theta+2x)=0 \\ \theta+2x=\frac{\pi}{2}+n \pi \\ 2x=\frac{\pi}{2}+n \pi-\theta \\ \text{ where } \theta=\arccos(\frac{2}{\sqrt{29}})$