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Author: Tinku Tara

Question-72291

Question Number 72291 by aliesam last updated on 27/Oct/19 Answered by mr W last updated on 27/Oct/19 $$\left(\mathrm{5}{m}\:\mathrm{sin}\:\left(\mathrm{3}{x}\right)\right)^{\mathrm{2}} =\left(\mathrm{4}\:\mathrm{sin}\:{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{5}{m}\:\mathrm{sin}\:\left(\mathrm{3}{x}\right)=\pm\mathrm{4}\:\mathrm{sin}\:{x} \\ $$$$\mathrm{5}{m}\:\mathrm{sin}\:{x}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right)=\pm\mathrm{4}\:\mathrm{sin}\:{x} \\…

lim-x-pi-2-1-tan-x-2-1-sin-x-1-tan-x-2-pi-2x-3-

Question Number 137820 by bramlexs22 last updated on 07/Apr/21 $$\underset{{x}\rightarrow\pi/\mathrm{2}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\left(\mathrm{1}+\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\left(\pi−\mathrm{2}{x}\right)^{\mathrm{3}} }\:? \\ $$ Answered by SLVR last updated on 07/Apr/21 Commented by SLVR last…

Question-6750

Question Number 6750 by Tawakalitu. last updated on 22/Jul/16 Answered by Yozzii last updated on 22/Jul/16 $${Let}\:\angle{QPR}=\angle{RPS}=\alpha>\mathrm{0},\:\angle{PQR}=\beta>\mathrm{0}\:{and}\:\angle{PSR}=\delta>\mathrm{0}. \\ $$$${By}\:{the}\:{sine}\:{rule},\:{in}\:\bigtriangleup{QPR},\: \\ $$$$\frac{{QR}}{{sin}\alpha}=\frac{{QP}}{{sin}\angle{PRQ}}=\frac{\mathrm{3}{a}}{{sin}\left(\pi−\left(\alpha+\beta\right)\right)} \\ $$$$\Rightarrow{QR}=\frac{\mathrm{3}{asin}\alpha}{{sin}\left(\alpha+\beta\right)}. \\ $$$${In}\:\bigtriangleup{PRS},\:{by}\:{the}\:{sine}\:{rule}\:{we}\:{have}…

Let-f-x-x-2-2x-3-x-1-amp-g-x-1-x-4-x-4-then-the-number-of-real-solutions-of-equation-f-x-g-x-is-

Question Number 137817 by bramlexs22 last updated on 07/Apr/21 $${Let}\:{f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3};\:{x}\geqslant\mathrm{1}\:\& \\ $$$${g}\left({x}\right)=\mathrm{1}+\sqrt{{x}+\mathrm{4}}\:;\:{x}\geqslant−\mathrm{4}\:{then} \\ $$$${the}\:{number}\:{of}\:{real}\:{solutions} \\ $$$${of}\:{equation}\:{f}\left({x}\right)={g}\left({x}\right)\:{is}\:… \\ $$ Answered by MJS_new last updated on…

cos-arctan-21-60-

Question Number 137813 by bramlexs22 last updated on 07/Apr/21 $$\mathrm{cos}\:\left(\mathrm{arctan}\:\left(\frac{\mathrm{21}}{\mathrm{60}}\right)\right)=? \\ $$ Answered by mr W last updated on 07/Apr/21 $${say}\:{t}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{21}}{\mathrm{60}} \\ $$$$\mathrm{tan}\:{t}=\frac{\mathrm{21}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{20}} \\…