Question

How do I find `x`: `2 sin x cos x-1=cos2x`?

I tried doing it:

`(2 sin x cos x)-1=2 cos^2x -1`
`(2 sin x cos x)-2 cos^2 x=0`

And I'm not sure if what I'm doing is correct?

Answer 1

`x=pi/2+2pin, x=(3pi)/2+2pin, x=pi/4+pin`, where `n` is any integer

Explanation:

Assuming the equation you want to solve is this:
`2sin(x)cos(x)-1=2cos^2(x)-1`

`2sin(x)cos(x)=2cos^2(x)`

`2sin(x)cos(x)-2cos^2(x)=0`

Then we factor it:
`2cos(x)(sin(x)-cos(x))=0`

For the expression to be equal to zero, either the first, second or both factors need to be equal to zero, so let's look at when the factors are equal to `0`:
`2cos(x)=0`

`cos(x)=0/2=0`

This equation actually has an infinite number of solutions since the `cos` function is periodic, so `x` can have these values, where `n` is any integer:
`x=pi/2+2pin, x=(3pi)/2+2pin`

Now, let's look at the zeroes for the other factor, `sin(x)-cos(x)`:
`sin(x)-cos(x)=0`

`(sin(x)-cos(x))/cos(x)=0/cos(x)`

`sin(x)/cos(x)-1=0`

`sin(x)/cos(x)=1`

Using the identity that `sin(x)/cos(x)=tan(x)`, we get:
`tan(x)=1`

This function is once again periodic, so the above equation has an infinite number of solutions as well. They are these, where `n` is any integer:
`x=pi/4+pin`

Combining these with the ones from the other equation, we get that these are the possible solutions to the original equation (of course, assuming the trigonometric arguments are in radians):
`x=pi/2+2pin, x=(3pi)/2+2pin, x=pi/4+pin`

Answer 2

`x=pi/2+kpi;k inZZ,x=pi/4+-2pik;k inZZ`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)cos2x=2cos^2x-1`

`rArr2sinxcosx-1=2cos^2x-1`

`rArr2sinxcosx-2cos^2x=0larrcolor(blue)"common factor of 2cosx"`

`rArr2cosx(sinx-cosx)=0`

`"equate each factor to zero and solve for x"`

`•color(white)(x)cosx=0rArrx=pi/2`

`"due to the periodicity of cos this value will repeat"`

`"general solution is "x=pi/2+-kpitok inZZ`

`•color(white)(x)sinx-cosx=0`

`rArrsinx=cosxrArrx=pi/4`

`"general solution is "x=pi/4+-2piktok inZZ`