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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

	$\frac{\tan x}{\tan y} = - 1 y = - x + k π (k \in Z)$ $k = 0 y = - x$ $k = 1 y = π - x$ $k = 0, 1 {\begin{cases} \sin x \cos x = \frac{1}{2} \Rightarrow x = \frac{π}{4} y = - \frac{π}{4} \\ \sin 2 x = - 1 \Rightarrow x = - \frac{π}{4} y = \frac{5 π}{4} \end{cases}$ $s o l u t i o n s :$ ${(- \frac{π}{4}, \frac{5 π}{4}); (\frac{π}{4}, - \frac{π}{4})}$
	Commented by MJS_new last updated on 27/Jan/23

	$‘ ‘ s o l v e i n R^{″}$ $I^{'} m afraid you have not found a l l solutions$
	Answered by a.lgnaoui last updated on 28/Jan/23
	$auther methode cos y=(1/(2sin x)) (1) x≠kπ (√(1−sin^2 x ))(√(1−(1/(4sin^2 x )))) =−(1/2) (1−sin^2 x)(1−(1/(4sin^2 x)))=(1/4) sin^2 x=z (1−z)(1−(1/(4z)))=(1/4) z^2 −z+(1/4)=0 △=0 z=(1/2) sin^2 x=(1/2)⇒sin x=±((√2)/2) x={−(π/4),+(π/4),((3π)/4),((5π)/4)}mod(2π) 1•x=−(π/4)⇒y=cos^(−1) ((1/(2sin x)))= =cos^(−1) (−((√2)/2)) y={((3π)/4),((5π)/4)}mod(2π) ((3π)/4) exclu y=((5π)/4) 2•x=+(π/4)⇒y=cos^(−1) (((√2)/2)) y={−(π/4)} ; (π/4)exclu 3•x=((3π)/4) y=+(π/4)(mod2π) 4•x=((5π)/4) y=((3π)/4)mod(2π) Resume −−−−−−−−−−−− S=(−(π/4),((5π)/4));(+(π/4),−(π/4)); (((3π)/4),+(π/4));(((5π)/4),((3π)/4))$
	$a u t h e r m e t h o d e$ $\cos y = \frac{1}{2 \sin x} (1) x \neq k π$ $\sqrt{1 - \sin^{2} x} \sqrt{1 - \frac{1}{4 \sin^{2} x}} = - \frac{1}{2}$ $(1 - \sin^{2} x) (1 - \frac{1}{4 \sin^{2} x}) = \frac{1}{4}$ $\sin^{2} x = z$ $(1 - z) (1 - \frac{1}{4 z}) = \frac{1}{4}$ $z^{2} - z + \frac{1}{4} = 0$ $△ = 0 z = \frac{1}{2}$ $\sin^{2} x = \frac{1}{2} \Rightarrow \sin x = \pm \frac{\sqrt{2}}{2}$ $x = {- \frac{π}{4}, + \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}} m o d (2 π)$ $1 ∙ x = - \frac{π}{4} \Rightarrow y = \cos^{- 1} (\frac{1}{2 \sin x}) =$ $= \cos^{- 1} (- \frac{\sqrt{2}}{2})$ $y = {\frac{3 π}{4}, \frac{5 π}{4}} m o d (2 π)$ $\frac{3 π}{4} e x c l u y = \frac{5 π}{4}$ $2 ∙ x = + \frac{π}{4} \Rightarrow y = \cos^{- 1} (\frac{\sqrt{2}}{2})$ $y = {- \frac{π}{4}}; \frac{π}{4} e x c l u$ $3 ∙ x = \frac{3 π}{4} y = + \frac{π}{4} (m o d 2 π)$ $4 ∙ x = \frac{5 π}{4} y = \frac{3 π}{4} m o d (2 π)$ $R e s u m e$ $- - - - - - - - - - - -$ $S = (- \frac{π}{4}, \frac{5 π}{4}); (+ \frac{π}{4}, - \frac{π}{4});$ $(\frac{3 π}{4}, + \frac{π}{4}); (\frac{5 π}{4}, \frac{3 π}{4})$
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