Question Number 17151 by Tinkutara last updated on 01/Jul/17 $$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta\:−\:\mathrm{2cos}\theta\:=\:\mathrm{4sin}\theta\:−\:\mathrm{sin2}\theta\:\mathrm{where} \\ $$$$\theta\:\in\:\left[\mathrm{0},\:\pi\right]\:\mathrm{is} \\ $$ Answered by ajfour last updated on 01/Jul/17 $$\mathrm{4sin}\:\theta−\mathrm{2sin}\:\theta\mathrm{cos}\:\theta+\mathrm{2cos}\:\theta−\mathrm{cos}\:^{\mathrm{2}}…
Question Number 17148 by Tinkutara last updated on 01/Jul/17 $$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cos}\:\left(\pi\sqrt{{x}\:−\:\mathrm{4}}\right)\:\mathrm{cos}\:\left(\pi\sqrt{{x}}\right)\:=\:\mathrm{1}\:\mathrm{is} \\ $$ Answered by ajfour last updated on 01/Jul/17 $$\Rightarrow\:\:\:\mathrm{cos}\:\left(\pi\sqrt{\mathrm{x}−\mathrm{4}}\right)=\pm\mathrm{1} \\ $$$$\mathrm{and}\:\:\:\mathrm{cos}\:\left(\pi\sqrt{\mathrm{x}}\right)=\pm\mathrm{1} \\…
Question Number 82654 by jagoll last updated on 23/Feb/20 $$\boldsymbol{\mathrm{I}}\mathrm{f}\:\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\mathrm{find}\:\mathrm{sin}\:\mathrm{x}+\:\mathrm{cos}\:\mathrm{x}\:? \\ $$ Commented by jagoll last updated on 23/Feb/20 Answered by mr W…
Question Number 17119 by gourav~ last updated on 01/Jul/17 Commented by prakash jain last updated on 01/Jul/17 $$\frac{\mathrm{sin}\:\left({A}+\mathrm{3}{B}\right)+\mathrm{sin}\:\left(\mathrm{3}{A}+{B}\right)}{\mathrm{sin}\:\mathrm{2}{A}+\mathrm{sin}\:\mathrm{2}{B}} \\ $$$$=\frac{\mathrm{2sin}\:\left(\frac{\mathrm{4}{A}+\mathrm{4}{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}{B}−\mathrm{2}{A}}{\mathrm{2}}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{2sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)\mathrm{cos}\:\left({A}−{B}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\…
Question Number 82638 by jagoll last updated on 23/Feb/20 $${given}\:{x}=\:\mathrm{cos}\:^{\mathrm{3}} {x} \\ $$$${what}\:{is}\:{x}\:? \\ $$ Commented by mr W last updated on 23/Feb/20 $${if}\:{you}\:{want}\:{to}\:{calculate}\:{it}\:{by}\:{hand}\:{by} \\…
Question Number 17100 by virus last updated on 30/Jun/17 $$\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 $$\mathrm{cos}\:\mathrm{5}{x}\:+\:\mathrm{cos}\:{x}\:=\:\mathrm{0} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\mathrm{3}{x}\:\mathrm{cos}\:\mathrm{2}{x}\:=\:\mathrm{0} \\…
Question Number 17095 by Tinkutara last updated on 30/Jun/17 $$\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 $${tanx}+{secx}=\mathrm{2}\Leftrightarrow{tanx}+\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}}=\mathrm{2}…
Question Number 17096 by Tinkutara last updated on 30/Jun/17 $$\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 17080 by Kunal kumar shukla last updated on 30/Jun/17 $$\mathrm{sin}^{\mathrm{4}} \theta/\mathrm{2}+\mathrm{cos}\:^{\mathrm{4}} \theta/\mathrm{2}\geqslant\mathrm{1}/\mathrm{2} \\ $$ Answered by virus last updated on 30/Jun/17 $$\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\theta/\mathrm{2}\right)\mathrm{cos}\:^{\mathrm{2}}…
Question Number 17073 by Kunal kumar shukla last updated on 30/Jun/17 Commented by Kunal kumar shukla last updated on 30/Jun/17 $${wrong}\:{answer} \\ $$ Commented by…