sin2x + 2sinx - cosx - 1 = 0

We know that:

sin2x = 2sinx*cosx

Let us substitute:

==> 2sinx*cosx + 2sinx - cosx - 1 = 0

Now factor 2sinx:

==> 2sinx( cosx +1 ) - (cosx + 1) = 0

Now factor (cosx + 1):

==> (cosx+1) *(...

## Get

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sin2x + 2sinx - cosx - 1 = 0

We know that:

sin2x = 2sinx*cosx

Let us substitute:

==> 2sinx*cosx + 2sinx - cosx - 1 = 0

Now factor 2sinx:

==> 2sinx( cosx +1 ) - (cosx + 1) = 0

Now factor (cosx + 1):

==> (cosx+1) *( 2sinx - 1 ) = 0

Then we have two options:

cosx + 1 = 0 OR 2sinx - 1 = 0

==> cosx = -1 OR sinx = 1/2

==> x1= pi + 2kpi

==> x2= pi/6 + 2kpi

==> x3= 5pi/6 + 2kpi

Then the answer is:

**x= { pi+ 2kpi , pi/6 + 2kpi , 5pi/6 + 2kpi}**