Find the number of solution(s) of following equations:
(i) `x^(2)-sinx-cosx+2=0` (ii) `sin^(2)x - 2sinx - x^(2)-3=0, x in [0,2pi]`

Explanation

We have,
(i) `x^(2)+2= sinx + cosx`
The range of `x^(2) + 2` is `[2, infty]`
and the range of `sinx + cosx` is `[-sqrt(2), sqrt(2)]`
As there is no common value ini the ranges hence no solution exists. (iii) `cos(x^(2) + x+1) = 2^(x) + 2^(-x)`
The range of `cos(x^(2)+x+1)` is `[-1,1]`
Using A.M. `ge` G.M.
`(2^(x) + 2^(-x))/(2) ge sqrt(2^(x). 2^(-x))`
`rArr r^(x) + 2^(-x) ge 2`
Hence, RHS `int [2, infty]`
Since, there is no common value in the ranges, hence number of solutions is zero.
(iii) The equation can be written as
`(sinx-1)^(2)=1`
Hence no solution exists.
(iv) `sin (pix)/(2sqrt(2))= (x-sqrt(2))^(2)+1`
LHS `int [-1,1]` and RHS `int [1,infty]`
Hence, solution is obtained at LHS = RHS=1
If (`x-sqrt(2))^(2) + 1=1 rArr x=sqrt(2)`
Also at `x=sqrt(2)`, LHS `=sin(pi xx sqrt(2))/(2sqrt(2))=1`
Hence, `x=1` is the only value satisfying the given equation and consequently number of solution is one.