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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.

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  • $\begingroup$ 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
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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 $360^\circ - 106.1^\circ = 253.9^\circ$

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  • $\begingroup$ 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
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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

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    $\begingroup$ 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

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