<s>[INST] Find $\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$ 
Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$
  without L'Hôpital's rule.

My work:
1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
2) Let $x=t-\frac{\pi}{6}$. Then 
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{3t}{\sin3t}\cdot\frac{1-2\sin \left(t-\frac{\pi}6\right)}{3t}$$
 [/INST] Hint: Multiply numerator and denominator by $$1+2\sin(x)$$
you can write your term in the form $$\frac{\csc \left(\frac{\pi }{4}-\frac{x}{2}\right) \csc
   \left(\frac{x}{2}+\frac{\pi }{4}\right)}{2 (2 \sin
   (x)+1)}$$
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