Why did I get extra solutions to $6\cos \theta + 7\sin \theta = 4 $?

Question

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 $$ \sin \theta = \frac{28\pm6\sqrt{69}}{85} $$ $$ \theta = 66.3^{\circ}, 113.7^{\circ}, 194.9^{\circ}, 345.1^{\circ} $$

Answer 1

Notice the equation you were trying to solve. That is: $$6\sqrt{1-x^2}=4-7x$$ A graph of such equation would look like this:

Which means, it only has one solution. When you square both sides of the equation, you're introducing one extra "strange root". That means you end up having an additional "fake" solution. Basically, now you have to check which of the solutions is a strange root. In this case, the correct answer is $$\sin {\theta}=\frac{28-6\sqrt{69}}{85}$$ which gives $$\theta \approx -15.07$$ From your solutions, only the values $113. 7°, 345.1°$ correspond to the original equation.

Answer 2

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.

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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.

$$\theta=345.1^\circ$$