what is the value of x in for which 2sin(1/2x)-cos(2x)=1?

If the first term is sin(x/2), then we'll use the half
angle identity:

sin(x/2) = sqrt[(1-cos
x)/2]

We'll use the double angle identity for the 2nd
term:

cos 2x = 2(cos x)^2 -
1

2sqrt[(1-cos x)/2] - 2(cos x)^2 + 1 =
1

We'll eliminate 1 both
sides:

2sqrt[(1-cos x)/2] - 2(cos x)^2 =
0

We'll divide by
2:

sqrt[(1-cos x)/2] - (cos x)^2 =
0

We'll move (cos x)^2 to the
right:

sqrt[(1-cos x)/2] = (cos
x)^2

We'll raise to
square:

(1-cos x)/2 = (cos
x)^4

1-cosx - 2(cos x)^4 =
0

We'll write 2(cos x)^4 = (cos x)^4 + (cos
x)^4

(1 - (cos x)^4) - cos x(1 + (cos x)^3) =
0

(1-cos x)(1+cos x)((1 + cos x)^4) - - cos x(1+cos x)(1 +
cos x + (cos x)^2) = 0

(1 + cos x)[(1 - cos x)(1 + (cos
x)^4) - cos x(1 + cos x +  (cos x)^2)] = 0

1 + cos x =
0

cos x = -1

x =
pi

The solution of the equation in the
interval (0 , 2pi) is x = pi.