Question-175547

Question Number 175547 by daus last updated on 02/Sep/22

Answered by mr W last updated on 02/Sep/22

$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}−{x}={x}+\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{x}={k}\pi+\frac{\pi}{\mathrm{12}} \\ $$$${in}\:\left(−\pi,\pi\right):\:−\frac{\mathrm{11}\pi}{\mathrm{12}},\:\frac{\pi}{\mathrm{12}} \\ $$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by CElcedricjunior last updated on 02/Sep/22

$$\left.\boldsymbol{{s}}\left.\boldsymbol{{inx}}=\boldsymbol{{cos}}\left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}\right)\:\boldsymbol{{for}}\:\boldsymbol{{x}}\in\right]−\pi;\boldsymbol{\pi}\right] \\ $$$$\boldsymbol{{or}}\:\boldsymbol{{sinx}}=\boldsymbol{{cos}}\left(−\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right) \\ $$$$=>\boldsymbol{{cos}}\left(−\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{2}}\right)=\boldsymbol{{cos}}\left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}\right) \\ $$$$=>\begin{cases}{\frac{\boldsymbol{\pi}}{\mathrm{2}}−\boldsymbol{{x}}=\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{3}}+\mathrm{2}\boldsymbol{{k}\pi}}\\{\frac{\pi}{\mathrm{2}}−\boldsymbol{{x}}=−\boldsymbol{{x}}−\frac{\boldsymbol{\pi}}{\mathrm{3}}+\mathrm{2}\boldsymbol{{k}}}\end{cases}\boldsymbol{{k}}\in\mathbb{Z} \\ $$$$=>\begin{cases}{\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{12}}+\boldsymbol{{k}\pi}}\\{}\end{cases} \\ $$$$\boldsymbol{{S}}_{\left.\right]\left.−\boldsymbol{\pi};\boldsymbol{\pi}\right]} =\left\{−\frac{\mathrm{11}\boldsymbol{\pi}}{\mathrm{12}};\frac{\boldsymbol{\pi}}{\mathrm{12}}\right\} \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\: \\ $$$$\:…….{le}\:{celebre}\:{cedric}\:{junior}……. \\ $$$$ \\ $$$$ \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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