Question Number 69953 by Shamim last updated on 29/Sep/19 $$\mathrm{if}\:\mathrm{tan}\:\theta+\mathrm{sec}\:\theta=\:\sqrt{\mathrm{3}}\:\mathrm{than}\:\mathrm{find}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\underset{} {\:}\theta\:\mathrm{where}\:\mathrm{0}^{\mathrm{o}} \leqslant\theta\leqslant\mathrm{2}\pi. \\ $$ Commented by Shamim last updated on 29/Sep/19 $$\mathrm{ans}\:\mathrm{ki}\:\mathrm{only}\:\mathrm{x}=\frac{\pi}{\mathrm{6}}??? \\…
Question Number 135435 by physicstutes last updated on 13/Mar/21 $$\mathrm{solve}\:\mathrm{in}\:\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{180}°\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{sin}\:\mathrm{3}{x}\:+\:\mathrm{cos}\:{x}\:=\:\mathrm{0}\: \\ $$ Answered by mr W last updated on 13/Mar/21 $$\mathrm{cos}\:{x}=−\mathrm{sin}\:\mathrm{3}{x}=\mathrm{sin}\:\left(−\mathrm{3}{x}\right)=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{3}{x}\right) \\ $$$$\frac{\pi}{\mathrm{2}}+\mathrm{3}{x}=\mathrm{2}{k}\pi\pm{x}…
Question Number 135419 by liberty last updated on 13/Mar/21 $$\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}.\mathrm{cot}\:{x}\:+\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}−\mathrm{cot}\:{x}−\mathrm{1}\:=\:\mathrm{0} \\ $$ Answered by benjo_mathlover last updated on 13/Mar/21 $$\Rightarrow\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}\left(\mathrm{cot}\:{x}+\mathrm{1}\right)−\left(\mathrm{cot}\:{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x}−\mathrm{1}\right)\left(\mathrm{cot}\:{x}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\\{\mathrm{cot}\:{x}=−\mathrm{1}\Rightarrow\mathrm{tan}\:{x}=−\mathrm{1}}\end{cases} \\…
Question Number 69871 by Shamim last updated on 28/Sep/19 $$\mathrm{Here},\:\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}\:\:} =\:\mathrm{4}\sqrt{\mathrm{mn}\:}\mathrm{and}\:\mathrm{tan}\theta+\mathrm{sin}\theta=\:\mathrm{m} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{tan}\theta−\mathrm{sin}\theta=\:\mathrm{n}. \\ $$ Answered by MJS last updated on 29/Sep/19 $$\mathrm{first}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve} \\…
Question Number 4216 by Rasheed Soomro last updated on 02/Jan/16 $$\boldsymbol{\mathrm{D}}\mathrm{etermine}\:\mathrm{a}\:\mathrm{triplet}\:\left(\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\right)\:\mathrm{of}\:\mathrm{real}\:\mathrm{numbers}\: \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{relation}\:\boldsymbol{\mathrm{a}}^{−\mathrm{1}} +\boldsymbol{\mathrm{b}}^{−\mathrm{1}} =\boldsymbol{\mathrm{c}}^{−\mathrm{1}} \mathrm{and}\:\:\mathrm{satisfying}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{conditions} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}>\boldsymbol{\mathrm{c}}\:\wedge\:\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}>\boldsymbol{\mathrm{a}}\:\wedge\:\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}\:. \\ $$ Terms of Service…
Question Number 135204 by victoras last updated on 11/Mar/21 Answered by benjo_mathlover last updated on 11/Mar/21 $$\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:+\:\mathrm{81}^{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:=\:\mathrm{30} \\ $$$$\mathrm{let}\:\mathrm{81}^{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \:=\:\mathrm{z}\:…
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Question Number 135206 by victoras last updated on 11/Mar/21 Answered by Olaf last updated on 11/Mar/21 $$\mathrm{81}^{\mathrm{sin}^{\mathrm{2}} {x}} +\mathrm{81}^{\mathrm{cos}^{\mathrm{2}} {x}} \:=\:\mathrm{30} \\ $$$$\mathrm{81}^{\mathrm{sin}^{\mathrm{2}} {x}} +\frac{\mathrm{81}}{\mathrm{81}^{\mathrm{sin}^{\mathrm{2}}…
Question Number 69641 by Henri Boucatchou last updated on 26/Sep/19 $$\boldsymbol{{Solve}}\:\:\boldsymbol{{arctg}}\left(\mathrm{1}−\boldsymbol{{x}}\right)\:+\:\frac{\mathrm{1}}{\boldsymbol{{arcctg}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135111 by faysal last updated on 10/Mar/21 Answered by Ñï= last updated on 10/Mar/21 $$\mathrm{2cos}\:\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}} \theta=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}\right)=\mathrm{2cos}\:\mathrm{2}\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}}…