1\left|\sin x\right|^{\sin ^2x-\sin x-2 }<1
|sinx|\ \epsilon \ [0,1] \\and\ (sinx-2)(sinx+1) \ \leq 0
if (sinx-2)(sinx+1) = 0 and < 0 then we dont get any solution except when |sinx| is zero.
so x = npi ? ...carelessness.
*edit : No solution.
11 Answers
1
106the power is (sinx+1)(sinx-2) which is always negative
So the expression is 1(sinx)>0 as sinx < 0 so the term is > 1 always
only case left to check is when sinx=1 then the value is 1
So, null set
i hope u get wat i meant to write :P
1@eureka .nul set is not posible put x=pi/2. 1<1..
if u put x=3pi/2.then (-1)0=1<1.again not possible....so why isnt my answer right ??????
24@asish
(sinx+1)(sinx-2) which is always negative and > 1 in magnitude
is it really ???
(sinx+1)(sinx-2) >0 ..I think
106NOO.. .. !!!
sinx<1 so sinx+1>0 and sinx-2<-1
leave that >1 in magnitude thats wrong
1oh sorry i made a mistake in my solution
at x = npi |sinx| is zero but (sinx-2)(sinx+1) < 0
so |sinx|(sinx-2)(sinx+1) → ∞
so we dont have a solution at x = npi. As for the rest my reasoning in #6 holds correct.
hence no solution