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}**

To find x if sin2x+2sinx-cosx-1 = 0

We know that sin2x = sin(x+x) = 2sinxcosx.

Therefore substituting sin2x = 2sinxcosx in the given equation, sin2x+2sinx -cosx -1 = 0, we get:

2sinxcosx +2sinx -cosx -1 = 0.

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

(cosx-1)(2sinx-1) = 0.

Equating each factor to zero we get:

cosx-1 = 0 . Or 2sinx -1= 0

cosx -1 = 0 gives cosx = 1. So x = 2npi, n = 0,1,2,...

2sinx -1 = 0 gives sinx = 1/2 . Or x = npi + (-1)^n*pi/6, n = 0, 1,2,....

To determine the angle x, we'll have to solve the equation. We notice that the first term of the equation is the function sine of a double angle.

We'll apply the formula for the double angle:

sin 2a = sin (a+a)=sina*cosa + sina*cosa=2sina*cosa

We'll replace 2a by 2x and we'll get:

sin 2a = 2sin x*cos x

We'll re-write the equation:

2sin x*cos x + 2sinx - cosx - 1 = 0

We'll factorize by 2sin x the first 2 terms:

2sinx(cos x + 1) - (cos x + 1) = 0

We'll factorize by (cos x + 1):

(cos x + 1)(2sin x - 1) = 0

We'll set each factor as zero:

cos x + 1 = 0

We'll add -1 both sides:

cos x = -1

x = arccos (-1)

x = pi

2sin x - 1 = 0

We'll add 1 both sides:

2sin x = 1

sin x = 1/2

x = arcsin (1/2)

x = pi/6

x = 5pi/6

**The angle x has the following values: {pi/6 ; 5pi/6 ; pi}.**