Solve this one

\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.

the 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

@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 ??????

k.i think i have gone mad

@asish
(sinx+1)(sinx-2) which is always negative and > 1 in magnitude

is it really ???

(sinx+1)(sinx-2) >0 ..I think

NOO.. .. !!!

sinx<1 so sinx+1>0 and sinx-2<-1

leave that >1 in magnitude thats wrong

@asish
sinx<1 so sinx+1>0

how ??

oh 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