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Question

Find range
$$\dfrac {1}{8-3\sin x}$$


Solution

We know that $$-1\le \sin{x}\le 1$$
$$\Rightarrow -3\le 3\sin{x}\le 3$$
$$\Rightarrow 3\ge -3\sin{x}\ge -3$$ (multiplying by $$-3$$)
$$\Rightarrow 8+3\ge 8-3\sin{x}\ge 8-3$$
$$\Rightarrow 11\ge 8-3\sin{x}\ge 5$$
$$\Rightarrow \dfrac{1}{11}\le\dfrac{1}{8-3\sin{x}}\le\dfrac{1}{5}$$
$$\therefore$$ Range is $$\left[\dfrac{1}{11},\dfrac{1}{5}\right]$$

Applied Mathematics

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