Find the value of `a` for which the equation a `sin(x+pi/4)=sin2x+9` will have real solution.

Explanation

`a sin (x + (pi)/(4)) = sin 2x + 9`
`rArr (a)/(sqrt(2))(sin x + cos x) = (sin x + cos x)^(2) + 8`
Putting `sin x + cos x = t,` the equation becomes
`t^(2)-(a)/(sqrt(2))t + 8 = 0, where - sqrt(2)letlesqrt(2)`
Since the product of root is 8, both roots cannot lie in `(-sqrt(2),sqrt(2)).` Thus,
`f(-sqrt(2)f(sqrt(2)) lt0`
`rArr (2 + a + 8 ) (2 - a + 8) lt 0`
or `( a+ 10) (10 - a) lt 0`
or `a lt -10 and a gt 0`
`rArr |a|gt 10`