How can I solve this problem if x is in the real number system and the equation is: sin(x+pi/6)+cos(pi/3-x)=1

sin(x+pi/6) + cos(pi/3) -x) =
1

First we will use trigonometric identities to
solve.

We know that:

sin(a+b)
= sina*cosb + cosa*sinb

==> sin(x+pi/6)=
sinx*cos(pi/6) + cosx*sin(pi/6)

==> sin(x+pi/6)=
(sqrt3 /2 )*sinx + (1/2)*cosx......(1)

cos(a-b)= cosa*cosb
+ sina*sinb

==> cos(pi/3 -x)= cospi/3*cosx +
sinpi/3*sinx

==> cos(pi/3 -x) = (1/2) cosx + sqrt3
/2 * sinx............(2)

Now we will add (1) and
(2):

==> 2(sqr3/2)sinx + 2(1/2)cosx =
1

==> sqr3*sinx + cosx =
1

==> sqrt3 sinx = 1-
cosx

Square both
sides:

==> 3sin^2 x = 1 - 2cosx + cos^2
x

==> 3(1-cos^2 x) = 1- 2cosx + cos^2
x

==> 3 - 3cos^2 x = 1- 2cosx + cos^2
x

==> 4cos^2 x - 2cosx -2 =
0

==> 2cos^2 x - cosx -1 =
0

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

==> cosx = 1 ==> x = 0, pi,
2pi

==> 2cosx +1 = 0 ==> cosx = -1/2
==> x = 4pi/3 , 5pi/3

==> x = {
0, pi, 4pi/3, 5pi/3, 2pi }