Separable first order ODE involving tangent

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Homework Statement
Consider the DE ##\frac{\mathrm{d}x}{\mathrm{d}t}=t\tan x## with initial value ##x(0)=\frac\pi{6}##.
1. Find the solution.
2. Describe the region in which the solutions are defined.
Relevant Equations
A separable first order ODE is of the form ##x'=g(x)h(t)##.
By inspection, we see that ##x=k\pi## is a solution for ##k\in\mathbb Z##. Moreover, the equation implicitly assumes ##x\neq n\pi/2## for odd ##n\in\mathbb Z##, since ##\tan x## isn't defined there. So suppose ##x\neq k\pi##, i.e. ##\tan x\neq 0##, then rearranging and writing ##\tan x=\frac{\sin x}{\cos x}## we have $$\frac{x'\cos x}{\sin x}=\frac{\mathrm{d}}{\mathrm{d}t}\log(|\sin x|)=t.$$ Integrating and exponentiating, we obtain, $$|\sin x|=Ae^{\frac{t^2}{2}}$$ The initial condition implies ##A=1/2##. Now, how do I get rid of the absolute values here? Does it somehow follow from the initial condition that ##\sin x## has to be positive?
 
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\sin x = \pm A e^{t^2/2}
where A >0. There are two cases of x>0 and x<0 during the time after. The initial condition says ##x(0)=\pi/6 >0 ## Our solution is the former one.
\sin x = \frac{e^{t^2/2}}{2}
 
I think from (\log |\sin x|)&#039; = t the next step is <br /> \log |\sin x| = \log |A| + \tfrac12 t^2 and hence <br /> |\sin x| = |A|e^{\frac12 t^2}. It follows from this that \sin x and A have the same sign, so we can drop the absolute value signs.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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