Hello-Solve-in-pi-pi-2sinx-cosx-sinx-0-result-should-be-in-radian-please-help-me-

Question Number 76477 by mathocean1 last updated on 27/Dec/19

Hello Solve in

] - π; π [

\frac{2 \sin x}{c o s x - s i n x} ⩾ 0

result should be in radian

please help me \dots

Answered by Kunal12588 last updated on 27/Dec/19

\frac{2 s i n x}{c o s x - s i n x} = \frac{\sqrt{2} s i n x}{s i n (\frac{π}{4} - x)} ⩾ 0

\Rightarrow x \in [0, \frac{π}{4}) \cup (- \frac{3 π}{4}, - π]

Commented by mathocean1 last updated on 27/Dec/19

please can you show me how you

have done to have \frac{\sqrt{2} \sin x}{s i n (\frac{π}{4} - x)}

Commented by Kunal12588 last updated on 27/Dec/19

c o s x - s i n x = \sqrt{2} (s i n \frac{π}{4} c o s x - c o s \frac{π}{4} s i n x)

= \sqrt{2} s i n (\frac{π}{4} - x)

\frac{2 s i n x}{\sqrt{2} s i n (\frac{π}{4} - x)} = \frac{\sqrt{2} s i n x}{s i n (\frac{π}{4} - x)}

Commented by mathocean1 last updated on 27/Dec/19

thank you sir \dots

I want to know how have solve the

rest \frac{\sqrt{2} \sin x}{\sin (\frac{π}{4} - x)} ⩾ 0 \boldsymbol because \boldsymbol i \boldsymbol have

\boldsymbol never \boldsymbol solved \boldsymbol this \boldsymbol type \boldsymbol of

\boldsymbol exercise

Commented by Kunal12588 last updated on 27/Dec/19

w e l l I h a v e a l s o n o t s o l v e s u c h t y p e o f q u e s t i o n s

I j u s t t h o u g h t o f a n i d e a t h a t i f n u m e r a t o r

i s g r e a t e r t h a n 0 t h e n d e n o m i n a t o r s h o u l d

a l s o b e g r e a t e r t h a n 0 s i m i l a r l y, w h e n N_{r} < 0

t h e n D_{r} < 0 .

Commented by mathocean1 last updated on 27/Dec/19

thank you once more sir \dots may God

bless you \dots

By the way, i wish you a happy

new year!

Answered by MJS last updated on 27/Dec/19

(1) \sin x ⩾ 0 \land \cos x - \sin x > 0

\lor

(2) \sin x ⩽ 0 \land \cos x - \sin x < 0

(1) \sin x ⩾ 0 \Leftrightarrow 0 ⩽ x < π

\cos x - \sin x > 0 \Leftrightarrow - \frac{3 π}{4} < x < \frac{π}{4}

\Rightarrow 0 ⩽ x < \frac{π}{4}

(2) \sin x ⩽ 0 \Leftrightarrow - π < x ⩽ 0

\cos x - \sin x < 0 \Leftrightarrow - π < x < - \frac{3 π}{4} \lor \frac{π}{4} < x < π

\Rightarrow - π < x < - \frac{3 π}{4}

\Rightarrow - π < x < - \frac{3 π}{4} \lor 0 ⩽ x < \frac{π}{4}

Commented by mr W last updated on 28/Dec/19

x = π i s m i s s i n g .

x = π i s m i s s i n g .

Commented by MJS last updated on 28/Dec/19

he typed] - π; π [\Leftrightarrow - π < x < π

he typed] - π; π [\Leftrightarrow - π < x < π

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