Question-185794

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

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})