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I need to find the general solution to the equation

$$\sin(x) + \sqrt3\cos(x)=\sqrt2$$

So I went ahead and divided by $2$, thus getting the form

$$\cos(x-\frac{\pi}{6})=\cos(\frac{\pi}{4})$$

Thus the general solution to this would be $$x = 2n\pi \pm\frac{\pi}{4}+\frac{\pi}{6}$$

Which simplifies out to be,

$$x = 2n\pi +\frac{5\pi}{12}$$ $$ x = 2n\pi -\frac{\pi}{12}$$

But the answer doesn't have the 2nd solution as a solution to the given equation. Did I go wrong somewhere?

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    $\begingroup$ Your answer seems to be the correct one. For example $x=-\frac {\pi} {12}$ does satisfy the given equation. $\endgroup$ Mar 5 '20 at 6:39
  • $\begingroup$ You solution is correct. May be they skip the second one. $\endgroup$
    – nmasanta
    Mar 5 '20 at 6:43
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As Kavi Rama Murthy's comment indicates, you haven't done anything wrong that I can see. You can quite easily very that $x = 2n\pi - \frac{\pi}{12}$ is a solution (coming from using $\cos\left(-\frac{\pi}{4}\right)$ on the right), as well as the first one you specify of $x = 2n\pi + \frac{5\pi}{12}$ (coming from using $\cos\left(\frac{\pi}{4}\right)$ on the right). Thus, it seems the answer has an oversight.

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  • $\begingroup$ Yes I guess so, it's a multiple choice question and both were given as options and only one was given correct, hence my confusion, I'm really just starting out with this topic $\endgroup$
    – Techie5879
    Mar 5 '20 at 6:40
  • $\begingroup$ @Techie5879 Unless there's some stated restriction on what $x$ could be, they are both valid options, so it seems the multiple-choice test has a mistake in it. $\endgroup$ Mar 5 '20 at 6:42
  • $\begingroup$ Got it. And no, there are no restrictions $\endgroup$
    – Techie5879
    Mar 5 '20 at 6:45

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