1.

Find the domain of 1/1-2sinx ​

Answer» ALIGN="absmiddle" ALT="{\bold{\BLUE{\sf{Let\:that\:expression\:as\:f(x)}}}}" class="latex-formula" id="TexFormula1" src="https://tex.z-dn.net/?f=%7B%5Cbold%7B%5Cblue%7B%5Csf%7BLet%5C%3Athat%5C%3Aexpression%5C%3Aas%5C%3Af%28x%29%7D%7D%7D%7D" TITLE="{\bold{\blue{\sf{Let\:that\:expression\:as\:f(x)}}}}">

{\bold{→}}{\bold{\red{\sf{1-2sinx≠0}}}}

{\bold{→}}{\bold{\red{\sf{sinx≠\frac{1}{2}}}}}

{\bold{→}}{\bold{\red{\sf{x≠sin^{-1} (\frac{1}{2})}}}}

{\boxed{\red{\sf{Domain \: x∈R-{\frac{π}{6}}}}}}

{\bold{\underline{\green{Extra\: information:}}}}

- 1 \leqslant \sin(x ) \leqslant 1 \\ - 2 \leqslant 2 \sin(x) \leqslant 2 \\ - ( - 2) \geqslant - 2 \sin(x) \geqslant - 2 \\ 1 + 2 \geqslant 1 - 2 \sin(x) \geqslant 1 - 2 \ \\ \ 3 \geqslant 1 - 2 \sin(x) \geqslant - 1 \\ \frac{1}{3} \geqslant \frac{1}{1 - 2 \sin(x) } \geqslant - \frac{1}{1}

{\bold{\sf{Range\: f(x)∈[-1,\frac{1}{3}]}}}


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