Question Number 17100 by virus last updated on 30/Jun/17 $$\mathrm{sec}\:{x}\mathrm{cos}\:\mathrm{5}{x}+\mathrm{1}=\mathrm{0} \\ $$$${find}\:{number}\:{of}\:{solution} \\ $$ Answered by Tinkutara last updated on 01/Jul/17 $$\mathrm{cos}\:\mathrm{5}{x}\:+\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{0} \\…
Question Number 148171 by ajfour last updated on 25/Jul/21 Commented by ajfour last updated on 25/Jul/21 $${This}\:{bird}\:{caged}\:{inside}\:{this}\: \\ $$$${tetrahedron}\:{has}\:{to}\:{start}\:{from} \\ $$$${origin},\:{touch}\:{each}\:{centroidal} \\ $$$${points}\:{of}\:{the}\:{faces}\:{and}\:{get} \\ $$$${back}.\:{If}\:{tetrahedron}\:{is}\:{regular}…
Question Number 148170 by Skabetix last updated on 25/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 17096 by Tinkutara last updated on 30/Jun/17 $$\mathrm{The}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\mathrm{tan}\:\mathrm{3}{x}\:−\:\mathrm{tan}\:\mathrm{2}{x}\:−\:\mathrm{tan}\:\mathrm{3}{x}\:\mathrm{tan}\:\mathrm{2}{x}\:=\:\mathrm{1}\:\mathrm{in} \\ $$$$\left[\mathrm{0},\:\mathrm{2}\pi\right]\:\mathrm{is} \\ $$ Answered by sma3l2996 last updated on 30/Jun/17…
Question Number 17095 by Tinkutara last updated on 30/Jun/17 $$\mathrm{The}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{tan}\:{x}\:+\:\mathrm{sec}\:{x}\:=\:\mathrm{2}\:\mathrm{which}\:\mathrm{lie}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{interval}\:\left[\mathrm{0},\:\mathrm{2}\pi\right]\:\mathrm{is} \\ $$ Answered by sma3l2996 last updated on 30/Jun/17 $${tanx}+{secx}=\mathrm{2}\Leftrightarrow{tanx}+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}=\mathrm{2}…
Question Number 82631 by Learner-123 last updated on 23/Feb/20 $${Eight}\:{point}\:{charges}\:{of}\:\mathrm{1}{nC}\:{each}\:{are} \\ $$$${located}\:{at}\:{corners}\:{of}\:{the}\:{cube}\:{in}\:{free} \\ $$$${space}\:{that}\:{is}\:\mathrm{1}{m}\:{on}\:{a}\:{side}\:.\:{Find}\:\:\mid\boldsymbol{{E}}\mid \\ $$$${at}\:{the}\:{centre}\:{of}\:{an}\:{edge}. \\ $$$$\left({Assume}\:{origin}\:{to}\:{be}\:{centre}\:{of}\:{cube}\right). \\ $$ Commented by Learner-123 last updated…
Question Number 82628 by jagoll last updated on 23/Feb/20 $${If}\:{a},{b},\:{c}\:{are}\:{in}\:{Harmonic}\:{progression} \\ $$$${find}\:{the}\:{value}\:{of}\:\frac{{a}+{b}}{{b}−{a}}\:+\:\frac{{b}+{c}}{{b}−{c}}\:.\:? \\ $$ Commented by $@ty@m123 last updated on 23/Feb/20 $$\frac{\mathrm{1}}{{a}},\:\frac{\mathrm{1}}{{b}},\frac{\mathrm{1}}{{c}}\:{are}\:{in}\:{AP} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{b}}=\boldsymbol{{d}},\:{say} \\…
Question Number 17093 by 1234Hello last updated on 02/Jul/17 $$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{function} \\ $$$$\mathrm{satisfying}\:{f}\left({x}\right).{f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\:{f}\left({x}\right)\:+\:{f}\left(\frac{\mathrm{1}}{{x}}\right)\:; \\ $$$${x}\:\in\:{R}\:−\:\left\{\mathrm{0}\right\}\:\mathrm{and}\:{f}\left(\mathrm{3}\right)\:=\:\mathrm{28},\:\mathrm{then}\:{f}\left(\mathrm{4}\right)\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to} \\ $$ Commented by prakash jain last updated on…
Question Number 82627 by Learner-123 last updated on 23/Feb/20 $${A}\:\mathrm{20}{nC}\:{point}\:{charge}\:{is}\:{located}\:{at} \\ $$$${P}\left(\mathrm{2},\mathrm{4},−\mathrm{3}\right)\:{in}\:{free}\:{space}.\:\boldsymbol{{F}}{ind}\:{the}\:{locus} \\ $$$${of}\:{all}\:{points}\:{at}\:{which}\:{E}_{{r}} =\mathrm{1}{V}/{m}. \\ $$ Commented by Learner-123 last updated on 26/Feb/20 $$??…
Question Number 148157 by aliibrahim1 last updated on 25/Jul/21 Answered by puissant last updated on 26/Jul/21 $$\mathrm{pgcd}\left(\mathrm{a};\mathrm{b}\right)=\mathrm{pgcd}\left(\mathrm{a}−\mathrm{b};\mathrm{b}\right) \\ $$$$\Rightarrow\:\mathrm{pgcd}\left(\mathrm{2}^{\mathrm{a}} −\mathrm{1};\mathrm{2}^{\mathrm{b}} −\mathrm{1}\right)=\mathrm{pgcd}\left(\mathrm{2}^{\mathrm{a}} −\mathrm{2}^{\mathrm{b}} ;\mathrm{2}^{\mathrm{b}} −\mathrm{1}\right) \\…