Solve $\cos x + \sqrt 3\sin x = \frac{\sqrt 3 -1}{\sqrt 2}$

trigonometry

Solve $\cos x + \sqrt 3 \sin x = \frac{\sqrt 3 -1}{\sqrt 2}, -\pi \leq x \leq \pi$

I tried to solve this equation as follows:

$2 \cos (x-\frac{\pi}{3}) = \frac{\sqrt 3 -1}{\sqrt 2}$

$\implies \cos (x-\frac{\pi}{3}) = \frac{\sqrt 3 -1}{2\sqrt 2}$

$\implies \cos (x-\frac{\pi}{3}) = \cos (\frac{5 \pi}{12})$

$\implies x = \frac{5 \pi}{12}+\frac{\pi}{3} + 2k \pi$

So, one value of $x$ is $\frac{3 \pi}{4}$

It can't be $\frac{7 \pi}{4}$ because this is greater than $\pi$. But, $\frac{7 \pi}{4}$ is the same as $\frac{- 3 \pi}{4}$. So, I thought $\frac{- 3 \pi}{4}$ should be a solution. I also tried to get another set of values by the following:

$ x = \pi – \frac{5 \pi}{12}+\frac{\pi}{3} + 2k \pi$

To verify my solution, I graphed it in desmos. None of the solutions seem right? Where am I going wrong? In general, I am able to solve a trig equation till the last step. However, I seem to face difficulties in finding all $x$ that satisfy a given trig. equation. I would be thankful if anyone pointed me towards some useful resources that may help.

Best Answer

You typed the wrong equation. The denominator has an extra factor of $2$.

Your actual work was correct, until the last two steps. In general, note that $\sin(\pi-x) = \sin(x)$ and $\cos(-x) = \cos(x)$.

Going in reverse, this means:

  1. If $\sin(x) = y$, then $x = \arcsin(y)+2\pi k$ or $x = \pi-\arcsin(y)+2\pi k$.
  2. If $\cos(x) = y$, then $x = \pm\arccos(y)+2\pi k$

So, for your general solution here, you'll have $x = \pm\dfrac{5\pi}{12}+\dfrac{\pi}{3} + 2k \pi$. Notice that what you did would've been the right thing if you had arcsine, but that's not the case here.

Of course, the problem isn't asking for a general solution for $k \in \mathbb Z$, but this point is still important nonetheless. I also thought this might help, given that you said about these problems. If you remember those two points, the general solution will be simple to obtain, so all that's left is using the domain that you're given.

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