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Question Number 109489 by bobhans last updated on 24/Aug/20
(1) sin (2x)−cos (2x)−sin (x)+cos (x)=0 (2)lim_(x→0) ((e^x −e^(−x) )/(sin x)) (3)lim_(x→−1) ∣x+1∣ sin (x+1)
$$\left(\mathrm{1}\right)\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{sin}\:{x}} \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\mid{x}+\mathrm{1}\mid\:\mathrm{sin}\:\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$
Answered by john santu last updated on 24/Aug/20
(2)lim_(x→0) ((e^x +e^(−x) )/(cos x)) = (2/1) = 2 (3) lim_(x→−1) ∣x+1∣sin (x+1)=0×0=0
$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{cos}\:{x}}\:=\:\frac{\mathrm{2}}{\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\mid{x}+\mathrm{1}\mid\mathrm{sin}\:\left({x}+\mathrm{1}\right)=\mathrm{0}×\mathrm{0}=\mathrm{0} \\ $$
Answered by john santu last updated on 24/Aug/20
(1)sin (2x)−sin (x)=cos (2x)−cos (x) ⇒2cos (((3x)/2))sin ((x/2))=−2sin (((3x)/2))sin ((x/2)) ⇒2sin ((x/2)){cos (((3x)/2))+sin (((3x)/2))}=0 { ((2sin ((x/2))=0⇒(x/2)=2kπ ; x = 4kπ)),((tan (((3x)/2))=−1⇒((3x)/2)=−(π/4)+k.π ; 3x=−(π/2)+2k.π)) :} ⇒ x =− (π/6)+((2k)/3).π
$$\left(\mathrm{1}\right)\mathrm{sin}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\left({x}\right)=\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left({x}\right) \\ $$$$\Rightarrow\mathrm{2cos}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)=−\mathrm{2sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{2sin}\:\left(\frac{{x}}{\mathrm{2}}\right)\left\{\mathrm{cos}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)+\mathrm{sin}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)\right\}=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{2sin}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{0}\Rightarrow\frac{{x}}{\mathrm{2}}=\mathrm{2}{k}\pi\:;\:{x}\:=\:\mathrm{4}{k}\pi}\\{\mathrm{tan}\:\left(\frac{\mathrm{3}{x}}{\mathrm{2}}\right)=−\mathrm{1}\Rightarrow\frac{\mathrm{3}{x}}{\mathrm{2}}=−\frac{\pi}{\mathrm{4}}+{k}.\pi\:;\:\mathrm{3}{x}=−\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}.\pi}\end{cases} \\ $$$$\Rightarrow\:{x}\:=−\:\frac{\pi}{\mathrm{6}}+\frac{\mathrm{2}{k}}{\mathrm{3}}.\pi \\ $$

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