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Question Number 185794 by Rupesh123 last updated on 27/Jan/23

Commented by Rupesh123 last updated on 27/Jan/23

Solve in R

Answered by a.lgnaoui last updated on 27/Jan/23

((tan x)/(tan y))=−1 y=−x+kπ (k∈Z) k=0 y=−x k=1 y=π−x k=0,1 { ((sin xcos x=(1/2) ⇒ x=(π/4) y=−(π/4) )),((sin2 x=−1 ⇒x=−(π/4) y=((5π)/4) )) :} solutions: {(−(π/4),((5π)/4));((π/4),−(π/4))}

$$\frac{\mathrm{tan}\:{x}}{\mathrm{tan}\:{y}}=−\mathrm{1}\:\:\:\:\:{y}=−{x}+{k}\pi\:\:\:\:\left({k}\in\mathbb{Z}\right) \\ $$$${k}=\mathrm{0}\:\:\:\:{y}=−{x} \\ $$$${k}=\mathrm{1}\:\:{y}=\pi−{x} \\ $$$${k}=\mathrm{0},\mathrm{1}\:\:\:\:\begin{cases}{\mathrm{sin}\:{x}\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\Rightarrow\:\:\:\:{x}=\frac{\pi}{\mathrm{4}}\:{y}=−\frac{\pi}{\mathrm{4}}\:}\\{\mathrm{sin2}\:{x}=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{x}=−\frac{\pi}{\mathrm{4}}\:\:\:\:\:\:{y}=\frac{\mathrm{5}\pi}{\mathrm{4}}\:\:}\end{cases}\: \\ $$$${solutions}: \\ $$$$\left\{\left(−\frac{\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}}\right);\left(\frac{\pi}{\mathrm{4}},−\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$

Commented by MJS_new last updated on 27/Jan/23

“solve in R” I′m afraid you have not found all solutions

$$``{solve}\:{in}\:\mathbb{R}'' \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{afraid}\:\mathrm{you}\:\mathrm{have}\:\mathrm{not}\:\mathrm{found}\:{all}\:\mathrm{solutions} \\ $$

Answered by a.lgnaoui last updated on 28/Jan/23

auther methode cos y=(1/(2sin x)) (1) x≠kπ (√(1−sin^2 x ))(√(1−(1/(4sin^2 x )))) =−(1/2) (1−sin^2 x)(1−(1/(4sin^2 x)))=(1/4) sin^2 x=z (1−z)(1−(1/(4z)))=(1/4) z^2 −z+(1/4)=0 △=0 z=(1/2) sin^2 x=(1/2)⇒sin x=±((√2)/2) x={−(π/4),+(π/4),((3π)/4),((5π)/4)}mod(2π) 1•x=−(π/4)⇒y=cos^(−1) ((1/(2sin x)))= =cos^(−1) (−((√2)/2)) y={((3π)/4),((5π)/4)}mod(2π) ((3π)/4) exclu y=((5π)/4) 2•x=+(π/4)⇒y=cos^(−1) (((√2)/2)) y={−(π/4)} ; (π/4)exclu 3•x=((3π)/4) y=+(π/4)(mod2π) 4•x=((5π)/4) y=((3π)/4)mod(2π) Resume −−−−−−−−−−−− S=(−(π/4),((5π)/4));(+(π/4),−(π/4)); (((3π)/4),+(π/4));(((5π)/4),((3π)/4))

$${auther}\:{methode} \\ $$$$\mathrm{cos}\:{y}=\frac{\mathrm{1}}{\mathrm{2sin}\:\:{x}}\:\:\:\:\left(\mathrm{1}\right)\:\:\:{x}\neq{k}\pi \\ $$$$\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\:}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4sin}\:^{\mathrm{2}} {x}\:}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4sin}\:^{\mathrm{2}} {x}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}={z} \\ $$$$\left(\mathrm{1}−{z}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{z}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${z}^{\mathrm{2}} −{z}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\bigtriangleup=\mathrm{0}\:\:\:\:{z}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} {x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sin}\:{x}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${x}=\left\{−\frac{\pi}{\mathrm{4}},+\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}}\right\}{mod}\left(\mathrm{2}\pi\right) \\ $$$$\mathrm{1}\bullet{x}=−\frac{\pi}{\mathrm{4}}\Rightarrow{y}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2sin}\:{x}}\right)= \\ $$$$=\mathrm{cos}^{−\mathrm{1}} \left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$\:\:{y}=\left\{\frac{\mathrm{3}\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}}\right\}{mod}\left(\mathrm{2}\pi\right) \\ $$$$\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:{exclu}\:\:\:{y}=\frac{\mathrm{5}\pi}{\mathrm{4}} \\ $$$$\mathrm{2}\bullet{x}=+\frac{\pi}{\mathrm{4}}\Rightarrow{y}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:{y}=\left\{−\frac{\pi}{\mathrm{4}}\right\}\:;\:\frac{\pi}{\mathrm{4}}{exclu} \\ $$$$\mathrm{3}\bullet{x}=\frac{\mathrm{3}\pi}{\mathrm{4}}\:\:\:\:{y}=+\frac{\pi}{\mathrm{4}}\left({mod}\mathrm{2}\pi\right) \\ $$$$\mathrm{4}\bullet{x}=\frac{\mathrm{5}\pi}{\mathrm{4}}\:\:\:{y}=\frac{\mathrm{3}\pi}{\mathrm{4}}{mod}\left(\mathrm{2}\pi\right) \\ $$$${Resume} \\ $$$$−−−−−−−−−−−− \\ $$$${S}=\left(−\frac{\pi}{\mathrm{4}},\frac{\mathrm{5}\pi}{\mathrm{4}}\right);\left(+\frac{\pi}{\mathrm{4}},−\frac{\pi}{\mathrm{4}}\right); \\ $$$$\left(\frac{\mathrm{3}\pi}{\mathrm{4}},+\frac{\pi}{\mathrm{4}}\right);\left(\frac{\mathrm{5}\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}}\right) \\ $$

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