Why did I get extra solutions to $6\cos \theta + 7\sin \theta = 4 $? 
Q. Find the values of $ \theta $ between $ 0^{\circ}$ and $360^{\circ}$ which satisfy the equation
$$ 6\cos \theta + 7\sin \theta = 4 $$

I solved this question differently to how I normally solve these questions, just to experiment and I ended up with two extra solutions. Please explain why this was so.
Workings:
Let $ \sin \theta = \frac{x}{1} $ then $ \cos \theta = \frac{\sqrt{1-x^2}}{1} $
$$6\times \sqrt{1-x^2} = 4-7x $$
$$ (6\times \sqrt{1-x^2})^2 = (4-7x)^2 $$
$$36 - 36x^2 = 16-56x + 49x^2 $$
$$85x^2 - 56x - 20 = 0 $$
$$ x = \frac{28\pm6\sqrt{69}}{85}$$
So $$ \sin \theta = \frac{28\pm6\sqrt{69}}{85} $$
$$ \theta = 66.3^{\circ}, 113.7^{\circ}, 194.9^{\circ}, 345.1^{\circ} $$

There are two mistakes.
(1) $\cos\theta$ can be negative. It can be $-\sqrt{1-x^2}$.
(2) Squaring an equation may lead to extra answers. But this mistake is neutralized by the first one.

The correct version should be:
Let $\displaystyle \sin \theta = \frac{x}{1} $ then $\displaystyle \cos \theta = \pm\frac{\sqrt{1-x^2}}{1} $
$$6\times \pm\sqrt{1-x^2} = 4-7x $$
$$ (6\times \pm\sqrt{1-x^2})^2 = (4-7x)^2 $$
$$36 - 36x^2 = 16-56x + 49x^2 $$
$$85x^2 - 56x - 20 = 0 $$
$$ x = \frac{28\pm6\sqrt{69}}{85}$$
When $\displaystyle x=\frac{28+6\sqrt{69}}{85}$, $x>0$ and $4-7x<0$. So we should take negative square root $-\sqrt{1-x^2}$ on L.H.S. Now $\sin\theta>0$ and $\cos\theta<0$. $\theta $ is in Quadrant II. 
$$\theta=113.7^\circ$$
When $\displaystyle x=\frac{28-6\sqrt{69}}{85}$, $x<0$ and $4-7x>0$. So we should take positive square root $\sqrt{1-x^2}$ on L.H.S. Now $\sin\theta<0$ and $\cos\theta>0$. $\theta $ is in Quadrant IV. 
$$\theta=345.1^\circ$$
