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Question Number 107304 by bemath last updated on 10/Aug/20

⋎bemath⋎ (1)Find domain of function f(x)= (√(log _(0.2) (((x+2)/(x−1)))−1)) (2) ((sin^2 (((9π)/8)−2x)−sin^2 (((7π)/8)−2x))/(sin (2018π+4x)))=?

$$\:\:\:\curlyvee{bemath}\curlyvee \\ $$$$\left(\mathrm{1}\right){Find}\:{domain}\:{of}\:{function}\: \\ $$$${f}\left({x}\right)=\:\sqrt{\mathrm{log}\:_{\mathrm{0}.\mathrm{2}} \left(\frac{{x}+\mathrm{2}}{{x}−\mathrm{1}}\right)−\mathrm{1}}\: \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{9}\pi}{\mathrm{8}}−\mathrm{2}{x}\right)−\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}−\mathrm{2}{x}\right)}{\mathrm{sin}\:\left(\mathrm{2018}\pi+\mathrm{4}{x}\right)}=? \\ $$

Answered by bobhans last updated on 10/Aug/20

✠bobhans✠ (1) D_f : log _(0.2) (((x+2)/(x−1)))−1≥ 0 log _(0.2) (((x+2)/(x−1))) ≥ log _(0.2) ((1/5)) ((x+2)/(x−1)) ≤ (1/5) ⇒((5x+10−x+1)/(5(x−1))) ≤0 ((4x+11)/(5(x−1))) ≤ 0 ⇒ −((11)/4)≤x<1 ...(1) ⇒((x+2)/(x−1)) > 0 ⇒ x < −2 ∪ x > 1 solution ⇒(1)∩(2) : x ≤ −((11)/4)

$$\:\:\:\maltese\boldsymbol{\mathrm{bobhans}}\maltese \\ $$$$\left(\mathrm{1}\right)\:\mathrm{D}_{\mathrm{f}} \::\:\mathrm{log}\:_{\mathrm{0}.\mathrm{2}} \left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{1}}\right)−\mathrm{1}\geqslant\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{log}\:_{\mathrm{0}.\mathrm{2}} \left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{1}}\right)\:\geqslant\:\mathrm{log}\:_{\mathrm{0}.\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{1}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow\frac{\mathrm{5x}+\mathrm{10}−\mathrm{x}+\mathrm{1}}{\mathrm{5}\left(\mathrm{x}−\mathrm{1}\right)}\:\leqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4x}+\mathrm{11}}{\mathrm{5}\left(\mathrm{x}−\mathrm{1}\right)}\:\leqslant\:\mathrm{0}\:\Rightarrow\:−\frac{\mathrm{11}}{\mathrm{4}}\leqslant\mathrm{x}<\mathrm{1}\:...\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}−\mathrm{1}}\:>\:\mathrm{0}\:\Rightarrow\:\mathrm{x}\:<\:−\mathrm{2}\:\cup\:\mathrm{x}\:>\:\mathrm{1} \\ $$$$\mathrm{solution}\:\Rightarrow\left(\mathrm{1}\right)\cap\left(\mathrm{2}\right)\::\:\mathrm{x}\:\leqslant\:−\frac{\mathrm{11}}{\mathrm{4}} \\ $$

Answered by bobhans last updated on 10/Aug/20

⌢Bobhans⌢ (2) (((sin (((9π)/8)−2x)+sin (((7π)/8)−2x))(sin (((9π)/8)−2x)−sin (((7π)/8)−2x)))/(sin 4x))= (((2sin (π−2x)cos ((π/8)))(2cos(π−2x)sin ((π/8))) )/(sin 4x))= ((sin ((π/4))(sin (2π−4x)))/(sin 4x))= ((((√2)/2) .sin 4x)/(sin 4x)) = ((√2)/2)

$$\:\:\:\:\:\:\:\:\:\smallfrown\mathcal{B}\mathrm{obhans}\smallfrown \\ $$$$\left(\mathrm{2}\right)\:\frac{\left(\mathrm{sin}\:\left(\frac{\mathrm{9}\pi}{\mathrm{8}}−\mathrm{2x}\right)+\mathrm{sin}\:\left(\frac{\mathrm{7}\pi}{\mathrm{8}}−\mathrm{2x}\right)\right)\left(\mathrm{sin}\:\left(\frac{\mathrm{9}\pi}{\mathrm{8}}−\mathrm{2x}\right)−\mathrm{sin}\:\left(\frac{\mathrm{7}\pi}{\mathrm{8}}−\mathrm{2x}\right)\right)}{\mathrm{sin}\:\mathrm{4x}}= \\ $$$$\frac{\left(\mathrm{2sin}\:\left(\pi−\mathrm{2x}\right)\mathrm{cos}\:\:\left(\frac{\pi}{\mathrm{8}}\right)\right)\left(\mathrm{2cos}\left(\pi−\mathrm{2x}\right)\mathrm{sin}\:\:\left(\frac{\pi}{\mathrm{8}}\right)\right)\:}{\mathrm{sin}\:\mathrm{4x}}= \\ $$$$\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}\right)\left(\mathrm{sin}\:\left(\mathrm{2}\pi−\mathrm{4x}\right)\right)}{\mathrm{sin}\:\mathrm{4x}}=\:\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:.\mathrm{sin}\:\mathrm{4x}}{\mathrm{sin}\:\mathrm{4x}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

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