The number of real numbers in [0,2pi] satisfying sin^(−1) x−2sin^2 x+1=0 is


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Question Number 79055 by gopikrishnan last updated on 22/Jan/20

$T h e n u m b e r o f r e a l n u m b e r s i n [0, 2 p i] s a t i s f y i n g \sin^{- 1} x - 2 {s i n}^{2} x + 1 = 0 i s$
Commented by mr W last updated on 22/Jan/20

$s i r : i a s k e d y o u r e p e a t e d l y n o t t o$ $w r i t e t h e w h o l e p o s t i n a s i n g l e l i n e .$ $b u t y o u s e e m t o i g n o r e t h a t a n d e v e n$ ${d o n}^{'} t g i v e a c o m m e n t . w h a t a p i t y!$ $n o w i j u s t a s k o t h e r m e m b e r s o f t h e$ $f o r u m n o t t o a n s w e r y o u r p o s t s .$
Commented by john santu last updated on 22/Jan/20

$i t s e e m s l i k e h e i g n o r e d i t$
Commented by mr W last updated on 23/Jan/20

$w h y a b s o l u t e l y n o c o m m u n i c a t i o n ?$ $m a y b e y o u h a v e r e a s o n s f o r t h a t w h a t$ $y o u d i d, t h e n t e l l u s! l e t u s u n d e r s t a n d$ $y o u!$ ${i t}^{'} s n o t m y p u r p o s e t o b o y c o t t y o u i n$ $t h i s f o r u m! w e s h o u l d o n l y r e s p e c t$ $e a c h o t h e r . i d o w h a t y o u r e q u e s t a n d$ $y o u d o w h a t i r e q u e s t . {t h a t}^{'} t h e r u l e!$
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