So I have an equation $2\sin2\theta=1-\cos\theta$ and my task was to find the solutions between the range of $0\leq\theta\leq360^\circ$ I can infer that $\cos\theta=-\frac{1}{2}$, and I realized something by substituting $\cos$ with $\sqrt{1-\sin^2\theta}$, I can infer that $\sin\theta=\frac{\sqrt{3}}{2}$.
Is this a viable method? Is my analysis correct?
On
A better way to test if your hypothesis is correct is to expand $2 \sin 2 \theta$, rearrange, and then substitute your values.
In your case, $4 \sin \theta \cos \theta + \cos \theta = 1 \rightarrow \cos \theta (4 \sin \theta + 1) = 1.$ Using the values you've supplied, $-\frac {1}{2} * 4 (\frac {\sqrt {3}}{2}) = -\sqrt {3},$ which is clearly not equal to $1.$
On the other hand, $2 \sin 2 \theta + \cos \theta = 1 \rightarrow \cos \theta (4 \sin \theta + 1) = 1$, so either $\cos \theta = 1$ or $\sin \theta = 0$. In either case, $\theta = 0°$ would be the only solution.
$$2\sin^2\dfrac\theta2=1-\cos\theta=2\sin2\theta$$
$$2\sin^2\dfrac\theta2=4\sin\dfrac\theta2\cos\dfrac\theta2\cos\theta$$
Either $\sin(\theta/2)=0,\theta=?$
Or $\sin(\theta/2)=2\cos\theta\cos(\theta/2)$
Or $\tan(\theta/2)=2\cos\theta$
Use https://www.cut-the-knot.org/arithmetic/algebra/WeierstrassSubstitution.shtml to form a cubic equation in $\tan(\theta/2)$