# 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

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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 }**