Find x if it is between -90 and 0 degrees and
$$\frac{\sqrt3}{\sin(x)} + \frac{1}{\cos(x)} = 4$$
I think I got the solution but I don't know how to close it out
= $$\sin(x) + \sqrt3\cos(x) = 4\sin(x)\cos(x)$$
By R-method and double angle
=$$ 2\sin(x+60) = 2sin(2x)$$
= $$\sin(x+60) = \sin(2x)$$
but I keep on getting confused on how to solve this such that it is inside a specific bound, is there a formula or something, especially if there are variables on each side.
On
HINT.-From your deduction $$\sin(x+60) = \sin(2x)\iff2\cos\left(\dfrac{3x+60}{2}\right)\sin\left(\dfrac{60-x}{2}\right)=0$$ Then because of $-90\le x\le 0$ you have the only solution $x=-\dfrac{4\pi}{9} (=-80^{\circ})$
On
I also like to see it in the following way. Recall cosine and sine are abscissa and ordinate of a point on the circle of radius $1$ centered in the origin. Thus you can rethink of your problem, as finding the intersection points between the line $$\lambda : \frac{\sqrt 3}{Y} + \frac{1}{X} = 4$$ and the circumference $$\gamma : X^2 + Y^2 = 1.$$
From the graph you readly see that the solutions are actually four. Let us find them trigonometrically.
Your final equation $$\sin\left(x + \frac{\pi}{3}\right) = \sin 2x$$ has the following sets of solutions.
First set is $$x + \frac{\pi}{3} = 2x + 2k \pi$$ yielding $$x = \frac{\pi}{3} + 2k \pi,$$ (red dot in the Figure) and second set is $$x + \frac{\pi}{3} = \pi - 2x + 2k\pi,$$ which gives $$x = -\frac{\pi}{9} + k \frac{2\pi}{3}$$ (blue dots in the Figure). The one you are interested in is of course in the fourth quadrant.
(Answer for the original question. For the edited question the answer is similar and I leave it to you to make appropriate changes) . Note that $|(x+\pi /3)- 2x| \leq \pi /3+\pi /2 =\pi$. Hence the only possibilites are $x+\pi /3=2x$ and $x+\pi /3=\pi - 2x$. This gives the solutions $x=\pi /3$ and $x=2\pi /9$.