`(cos2x + sinx)/sqrt(sin(x - pi/4)) = 0`
For this expression to be equal to zero, the numerator should be equal to zero. But given that the denomiantor is not also zero.
First let's check the denominator,
`sqrt(sin(x-pi/4))`
Therefore `sin(x-pi/4) != 0`
So x is not equal to `pi/4`
`x!=pi/4`
Now let's find the roots for this equation.
`cos(2x)+sin(x) = 0`
We can write `cos(2x)` in terms of `sin(x)` .
`cos(2x) = 1 -2sin^2(x)`
Therefore the equation changes into,
`1 -2sin^2(x)+sin(x) = 0`
Rearranging,
`2sin^2(x) -sin(x)-1 = 0`
By solving this quadratic equation, we can find solutions for sin(x).
`(2sin(x)+1)(sin(x)-1) = 0`
`sin(x) = -1/2` or `sin(x) = 1`
There are two solutions. Let's consider one by one.
`sin(x) = -1/2`
`x = -pi/6`
This is the primary solution. The general solution for sine is given by,
`x = npi+(-1)^n(-pi/6)` where n is any integer.
The other solution is, sin(x) = 1.
The primary solution is `x = pi/2`
The general solution is,
`x = npi+(-1)^n(pi/2)` where n is any integer.
Therefore the solutions are,
`x = npi+(-1)^n(-pi/6)` where n is any integer.
and
`x = npi+(-1)^n(pi/2)` where n is any integer.