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


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?

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.