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.


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

  • $\begingroup$ Squaring can often introduce extra unwanted solutions to equations. For example, the argument $x=-2 \implies x^2 = 4 \implies x = \pm 2$, giving spurious solutions. It's good practice to check your resulting solutions for $x$ work in the original equation. $\endgroup$ – B. Mehta Apr 9 '18 at 2:09

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.


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.


  • $\begingroup$ Sorry, it's really late and I'm being a bit slow. Please could you explain how noting that $ x < 0 $ and $4-7x > 0 $ leads you to take a positive square root $\endgroup$ – Omniscient Apr 9 '18 at 2:35
  • $\begingroup$ @Omniscient As $28^2<(6\sqrt{69})^2$, $\frac{28-6\sqrt{69}}{85}<0$. $\endgroup$ – CY Aries Apr 9 '18 at 2:38
  • $\begingroup$ We can find $4-7x$ when $x=\frac{28-6\sqrt{69}}{85}$. It is $\frac{144+42\sqrt{69}}{85}$ which is positive. $\endgroup$ – CY Aries Apr 9 '18 at 2:41
  • $\begingroup$ @Omniscient I have edited my answer and make it easier to read. $\endgroup$ – CY Aries Apr 9 '18 at 2:42

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:

enter image description here

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.

  • $\begingroup$ The answers should be $113.7^{\circ}$ and $345.1^{\circ}$. $\endgroup$ – CY Aries Apr 9 '18 at 2:24
  • $\begingroup$ Yeah, my bad! I forgot to consider the values of cosine. $\endgroup$ – NotAMathematician Apr 9 '18 at 17:05

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