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Question Number 131295 by EDWIN88 last updated on 03/Feb/21
$${Find}\:{domain}\:{of}\:{function}\: \\ $$$${f}\left({x}\right)=\:\sqrt{\frac{\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{6}+\mathrm{35}{x}−\mathrm{6}{x}^{\mathrm{2}} }}\: \\ $$
Answered by liberty last updated on 03/Feb/21
![f(x)=(√((cos x−1/2)/(6+35x−6x^2 ))) determinant (((cos x≥(1/2)),(6x^2 −35x−6<0)),((−(π/3)≤x≤(π/3) ∪ ((5π)/3)≤x≤((7π)/3)),((6x+1)(x−6)<0))) ∴ x ∈ (−(1/6),(π/3) ] ∪ [ ((5π)/3), 6)](Q131297.png)
$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{cos}\:\mathrm{x}−\mathrm{1}/\mathrm{2}}{\mathrm{6}+\mathrm{35x}−\mathrm{6x}^{\mathrm{2}} }}\: \\ $$$$\:\begin{array}{|c|c|}{\mathrm{cos}\:\mathrm{x}\geqslant\frac{\mathrm{1}}{\mathrm{2}}}&\hline{\mathrm{6x}^{\mathrm{2}} −\mathrm{35x}−\mathrm{6}<\mathrm{0}}\\{−\frac{\pi}{\mathrm{3}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{3}}\:\cup\:\frac{\mathrm{5}\pi}{\mathrm{3}}\leqslant\mathrm{x}\leqslant\frac{\mathrm{7}\pi}{\mathrm{3}}}&\hline{\left(\mathrm{6x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{6}\right)<\mathrm{0}}\\\hline\end{array} \\ $$$$\:\therefore\:\mathrm{x}\:\in\:\left(−\frac{\mathrm{1}}{\mathrm{6}},\frac{\pi}{\mathrm{3}}\:\right]\:\cup\:\left[\:\frac{\mathrm{5}\pi}{\mathrm{3}},\:\mathrm{6}\right)\: \\ $$