Solve $2\cos x^{\circ} + 3\sin x^{\circ} = -1$ where $0 \le x \le 360$

I am not asked to use any form so I am going to use $k\cos(x-alpha)$

$2\cos x^{\circ} + 3\sin x^{\circ} = k\cos(x-alpha)$

$$= k(\cos x\cos\alpha + \sin x\sin\alpha)$$

$$= k\cos\alpha \cos x + k\sin\alpha \sin x$$

Equating coefficients:

$k\cos\alpha = 2$

$k\sin\alpha = 3$

$\alpha$ is in the 1st quadrant because both sin and cos are positive.

$\alpha = \arctan \frac{3}2 = 56.3$

$k = \sqrt{3^2 + 2^2} = \sqrt{13}$

$\therefore \sqrt{13}\cos(x-56.3) = -1$

$(x-56.3) = \arccos\frac{-1}{\sqrt{13}}$

$x - 56.3 = 106.1$

$x = 162.4$

From here I am completely clueless of what to do. I cannot even find any links about this from my googling.

The answer to this question has 2 answers, 162.4 and 310.2 but solving this equation $3\cos x + 4\sin x = 5$ only has one answer for x which is mystifying and illogical to me.

There is more than one answer for x and I do not know why. I've seen this:

$\implies x-\alpha=360^\circ n\pm60^\circ$ where $n$ is any integer

But I have no idea what this means or what n is or where it comes from and do the 4 quadrants influence my answer?

Can anyone provide an idiots guide to how I can solve for x when I get to the last parts of these problems? Even a link to something that explains it would help.

  • $\begingroup$ Because you are using $x^\circ{}$ in the argument to trigonometric functions, the domain should be either $0^\circ{} \le x^\circ{} \le 360^\circ{}$ or $0 \le x \le 360$ but not $0^\circ{} \le x \le 360^\circ{}$. $\endgroup$ – lastresort Apr 24 '16 at 6:25

Below is a picture of the two places on the unit circle where $\cos \theta = -\dfrac{1}{\sqrt{13}}$.

enter image description here

The general solution, in degrees, would be $\theta = 360^\circ n \pm 106.1^\circ$ where $n \in \mathbb Z$

The two answers between $0^\circ$ and $360^\circ$ would be

$0^\circ + 106.1^\circ = 106.1^\circ$ and $360^\circ - 106.1^\circ = 253.9^\circ$

  • $\begingroup$ what is n, I have not read about this $\endgroup$ – dagda1 Apr 24 '16 at 13:04
  • $\begingroup$ I see now why cos is negative in second and third quadrants. I can rest at last :) $\endgroup$ – dagda1 Apr 24 '16 at 14:14
  • $\begingroup$ @dagda1 - $\cos$ has a period of $360^\circ$. This is a generalization of $\cos x = \cos(x + 360^\circ)$. $\endgroup$ – steven gregory Apr 25 '16 at 6:21

You actually made a mistake by assuming that alpha is postive because tan alpha is positive in both the first and third quadrants therefore omitting one solution

  • 1
    $\begingroup$ Why is it positive in the 3rd quadrant and why is it not positive in this equation $3\cos x+4\sin x=5$ which only has value for x $\endgroup$ – dagda1 Apr 24 '16 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.