Solve the equation \(\operatorname{cosec}^2 2x+\dfrac2{\tan2x}=1\) for \(-\dfrac\pi2\le x\le\dfrac\pi2\).[7]
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Answer:For the printed expression: \(x=-\dfrac12\tan^{-1}\dfrac12\approx-0.232\), or \(x=\dfrac12(\pi-\tan^{-1}\dfrac12)\approx1.34\). See domain note.