Your solution(s) appear wrong. One tip: there are often multiple ways to solve elementary trigonometric equations. Your first instinct might be to immediately apply a double angle formula and solve the resulting quadratic. While this should work, it is unnecessary work, and may lead to additional errors (as it probably has in your case). Instead, wherever possible, keep things linear and simple.
In this case, i would just rearrange to: $\cos 2x = -\frac 23$ and solve for that.
I'm going to start by letting $y = 2x$; it's not something you need to do when you're more experienced, but it's illustrative at this stage. We need to first find $y$ such that:
1) its cosine equals $-\frac 23$ (a negative number)
2) it lies in the correct range. Since $0 \leq x \leq 180^{\circ}, 0 \leq y \leq 360^{\circ}$
The range $0 \leq y \leq 360^{\circ}$ means we're considering one full "rotation" around the Cartesian plane. Note that angles are measured counter-clockwise from the positive direction of the horizontal axis. I hope you know how the four quadrants are named - counterclockwise around the plane.
Now, at this point, I can simply quote the periodic nature of the trigonometric ratios. However, I've found a mnemonic device actually helps solutions immensely. This simple mnemonic you should master is "ASTC" - All (1st), Sine (2nd), Tangent (3rd), Cosine (4th). Basically that means all ratios are positive in the first quadrant, only sine is positive in the second, and so forth. The ratios are negative in any quadrant where they are not specified to be positive.
From that, you can deduce that cosine is positive in the first (angles between $0$ and $90$ degrees) and fourth (angles between $270$ and $360$ degrees) quadrants only. It is negative in the second and third quadrants. This means we are looking for $y$ in the range (in degrees): $90^{\circ} \leq y \leq 270^{\circ}$.
At this point, pull out your calculator and find the reference angle by evaluating the inverse cosine of $\frac 23$. Note that you can ignore the sign here. You'll get $\arccos \frac 23 \approx 48.19^{\circ}$.
To get the required results, you need to translate that reference angle to the correct range in order to ensure the cosine is negative as required. To get an answer in the second quadrant, subtract it from $180^{\circ}$, i.e find $180^{\circ}- 48.19^{\circ} = 131.81^{\circ}$. To get an answer in the third quadrant, add $180^{\circ}$ to it, i.e. find $180^{\circ}+ 48.19^{\circ} = 228.19^{\circ}$. Those are the two values of $y$ you require.
The reason why you do this should become apparent if you sketch the coordinate axes. The transformations you apply to the reference angle change only the sign of the answer, but don't affect the numerical value. If you want an answer in the first quadrant, just take the reference angle. Second quadrant, subtract the reference angle from $180$ degrees. Third quadrant: add $180$ degrees to the reference angle. Fourth quadrant, subtract the reference angle from $360$ degrees. This will always give correct angles in the range $0$ to $360$ degrees.
I know it's not a very good diagram, but I hope it helps clarify things, along with what I've written above:
Since we defined $y=2x$, the values of $x$ are just half of those we got for $y$, and you get: $65.91^{\circ}$ and $114.09^{\circ}$, which you can verify by substituting back into the original equation.
Best Answer
I do one, you do the other:
$$\tan^22x=1\iff \tan 2x=\pm1\iff 2x=\pm\frac\pi4+k\pi\;,\;\;k\in\Bbb Z\iff$$
$$\iff x=\pm\frac\pi8+k\frac\pi2\;,\;\;k\in\Bbb Z$$
Hint for the other:
$$\sin3x=-\frac14\iff3x=\arcsin\left(-\frac14\right)+2k\pi\ldots\ldots\text{etc.}$$