Values of x and y satisfying the equation sin7y=|x3−x2−9x+9|+|x3−x2−4x+4|+sec22y+cos4y are
x=1,y=nπ,n∈I
x=1,y=2nπ+π2,n∈I
x=1,y=2nπ,n∈I
none of the above
sin7y=|x3−x2−9x+9|+|x3−x2−4x+4| +sec22y+cos4y
For x=1
sin7y=sec22y+cos4y
⇒sin7ycos22y=1+cos4ycos22y
Since, LHS≤1. and RHS≥1
Which is possible only when
∴sin7ycos22y=1
⇒sin7y=1 andcos22y=1;y=π2
General value of y is 2nπ+π2
Hence, x=1 and y=2nπ+π2,n∈I