solve for x

can u provide the actual question.

http://www.wolframalpha.com/input/?i=%28sinx%29^3%2B%28sinx%29^2%2Bsinx%3D1
check here- the answer is ridiculous !

how to solve?

sin³x+sin²x+sinx=1
(sin²x+1)(sinx+1)=2
0≤ sin²x+1≤2 and 0≤sinx+1≤2
now integral factors of 2 are not possible
therefore
sinx+1=2^p and sin²x+1=2^1-p
where p=[0,1]
hence
x=k(pi)+(-1)^karcsin(2^p-1) and x=l(pi)±arcsin(2^1-p-1)
where k,l are integers and p=[0,1]

hari shankar sir, nishant sir and nasiko sir please try,,,,,,

I think you should check the question once. Though in this case we shall get a single value of sin x, however that is not very pretty and I had to solve the cubic in order to get that value.