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Category: Algebra

Question-148706

Question Number 148706 by mathdanisur last updated on 30/Jul/21 Answered by mindispower last updated on 30/Jul/21 $${sin}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\ $$$$\left({E}\right)\Leftrightarrow\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{8}{x}\right)+{cos}\left(\mathrm{12}{x}\right)+{cos}\left(\mathrm{14}{x}\right)\right)=\mathrm{2} \\ $$$$\Leftrightarrow{cos}\left(\mathrm{6}{x}\right)+{cos}\left(\mathrm{14}{x}\right)+{cos}\left(\mathrm{8}{x}\right)+{cos}\left(\mathrm{12}{x}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}{cos}\left(\mathrm{10}{x}\right){cos}\left(\mathrm{4}{x}\right)+\mathrm{2}{cos}\left(\mathrm{10}{x}\right){cos}\left(\mathrm{2}{x}\right)=\mathrm{0} \\…

Solve-for-equation-2tg-3x-3tg-2x-tg-2-2x-tg-3x-

Question Number 148707 by mathdanisur last updated on 30/Jul/21 $${Solve}\:{for}\:{equation}: \\ $$$$\mathrm{2}{tg}\left(\mathrm{3}{x}\right)\:-\:\mathrm{3}{tg}\left(\mathrm{2}{x}\right)\:=\:{tg}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:\centerdot\:{tg}\left(\mathrm{3}{x}\right) \\ $$ Answered by nimnim last updated on 30/Jul/21 $${Let}\:{me}\:{give}\:{a}\:{try}…. \\ $$$$\Rightarrow\mathrm{2}\left(\frac{\mathrm{3}{tanx}−{tan}^{\mathrm{3}}…

lim-n-0-1-nx-1-n-2-x-4-dx-

Question Number 148655 by mathdanisur last updated on 29/Jul/21 $$\underset{\boldsymbol{{n}}\rightarrow\infty} {{lim}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{{nx}}{\mathrm{1}\:+\:{n}^{\mathrm{2}} {x}^{\mathrm{4}} }\:{dx}\:=\:? \\ $$ Answered by ArielVyny last updated on 29/Jul/21 $${n}\int_{\mathrm{0}}…

lim-x-1-1-x-2-x-1-x-1-x-

Question Number 148654 by mathdanisur last updated on 29/Jul/21 $$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)^{\frac{\mathrm{1}−\sqrt{\boldsymbol{{x}}}}{\mathrm{1}−\boldsymbol{{x}}}} \:=\:? \\ $$ Answered by ArielVyny last updated on 29/Jul/21 $${e}^{\frac{\mathrm{1}−\sqrt{{x}}}{\mathrm{1}−{x}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{2}+{x}}\right)} \:\:\:\:{posons}\:{t}={x}−\mathrm{1}\:\begin{cases}{{x}\rightarrow\mathrm{1}}\\{{t}\rightarrow\mathrm{0}}\end{cases} \\ $$$${e}^{\frac{\sqrt{{t}+\mathrm{1}}−\mathrm{1}}{{t}}{ln}\left(\frac{{t}+\mathrm{2}}{{t}+\mathrm{3}}\right)}…

Question-148630

Question Number 148630 by mathdanisur last updated on 29/Jul/21 Answered by mr W last updated on 30/Jul/21 $$\angle{C}=\mathrm{60}−\mathrm{20}=\mathrm{40}° \\ $$$$\angle{A}=\mathrm{180}−\mathrm{20}−\mathrm{60}=\mathrm{100}° \\ $$$${say}\:{AB}={a} \\ $$$$\frac{{AF}}{\mathrm{sin}\:\mathrm{20}}=\frac{{BF}}{\mathrm{sin}\:\mathrm{100}}=\frac{{AB}}{\mathrm{sin}\:\mathrm{60}}=\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}} \\…