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Question-175547




Question Number 175547 by daus last updated on 02/Sep/22
Answered by mr W last updated on 02/Sep/22
cos ((π/2)−x)=cos (x+(π/3))  ⇒2kπ+(π/2)−x=x+(π/3)  ⇒x=kπ+(π/(12))  in (−π,π): −((11π)/(12)), (π/(12))
$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}−{x}={x}+\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{x}={k}\pi+\frac{\pi}{\mathrm{12}} \\ $$$${in}\:\left(−\pi,\pi\right):\:−\frac{\mathrm{11}\pi}{\mathrm{12}},\:\frac{\pi}{\mathrm{12}} \\ $$
Commented by Tawa11 last updated on 15/Sep/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by CElcedricjunior last updated on 02/Sep/22
sinx=cos(x+(𝛑/3)) for x∈]−π;𝛑]  or sinx=cos(−x+(𝛑/2))  =>cos(−x+(𝛑/2))=cos(x+(𝛑/3))  => { (((𝛑/2)−x=x+(𝛑/3)+2k𝛑)),(((π/2)−x=−x−(𝛑/3)+2k)) :}k∈Z  => { ((x=(𝛑/(12))+k𝛑)),() :}  S_(]−𝛑;𝛑]) ={−((11𝛑)/(12));(𝛑/(12))}               .......le celebre cedric junior.......
$$\left.\boldsymbol{{s}}\left.\boldsymbol{{inx}}=\boldsymbol{{cos}}\left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}\right)\:\boldsymbol{{for}}\:\boldsymbol{{x}}\in\right]−\pi;\boldsymbol{\pi}\right] \\ $$$$\boldsymbol{{or}}\:\boldsymbol{{sinx}}=\boldsymbol{{cos}}\left(−\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right) \\ $$$$=>\boldsymbol{{cos}}\left(−\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)=\boldsymbol{{cos}}\left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}\right) \\ $$$$=>\begin{cases}{\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}=\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}+\mathrm{2}\boldsymbol{{k}\pi}}\\{\frac{\pi}{\mathrm{2}}−\boldsymbol{{x}}=−\boldsymbol{{x}}−\frac{\boldsymbol{\pi}}{\mathrm{3}}+\mathrm{2}\boldsymbol{{k}}}\end{cases}\boldsymbol{{k}}\in\mathbb{Z} \\ $$$$=>\begin{cases}{\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{12}}+\boldsymbol{{k}\pi}}\\{}\end{cases} \\ $$$$\boldsymbol{{S}}_{\left.\right]\left.−\boldsymbol{\pi};\boldsymbol{\pi}\right]} =\left\{−\frac{\mathrm{11}\boldsymbol{\pi}}{\mathrm{12}};\frac{\boldsymbol{\pi}}{\mathrm{12}}\right\} \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:…….{le}\:{celebre}\:{cedric}\:{junior}……. \\ $$$$ \\ $$$$ \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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