for-pi-lt-x-lt-pi-2-find-the-solution-of-sin-x-gt-sin-x-2-2-sin-x-1-

Question Number 131622 by rydasss last updated on 06/Feb/21

f o r - π < x < - \frac{π}{2} f i n d t h e s o l u t i o n o f

s i n x > \frac{s i n x + 2}{2 s i n x + 1}

Answered by mr W last updated on 07/Feb/21

- π < x < - \frac{π}{2} \Rightarrow - 1 < \sin x < 0

c a s e 1 : 2 \sin x + 1 > 0, i . e . \sin x > - \frac{1}{2}

\sin x (2 \sin x + 1) > \sin x + 2

\sin^{2} x > 1

\Rightarrow n o s o l u t i o n!

c a s e 2 : 2 \sin x + 1 < 0, i . e . \sin x < - \frac{1}{2}

\sin x (2 \sin x + 1) < \sin x + 2

\sin^{2} x < 1

\Rightarrow \sin x > - 1

\Rightarrow - \frac{5 π}{6} < x < - \frac{π}{2} (s o l u t i o n)

Commented by rydasss last updated on 07/Feb/21

T h a n k y o u s o m u c h .

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