`(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`

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

`(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.**