`sqrt(3)sinx+cosx=1`

`sqrt(3)sinx=1-cosx`

`3sin^2x=1-2cosx+cos^2x` -- square both sides

`3(1-cos^2x)=1-2cosx+cos^2x` -- Pythagorean relationship

`4cos^2x-2cosx-2=0` -- clear parantheses and collect

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

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

`cosx=-1/2` or `cosx=1`

** note that by squaring both sides, we might have introduced extraneous solutions, so we must check all potential solutions in the original equation.

`cosx=1 => x=0 +- 2pi`

This works as sin0=0 and cos0=1

`cosx=-1/2 => x=(2pi)/(3)` or `x=(4pi)/3`

However, `x=(4pi)/3` is not a solution to the original equation since the sin is negative in the third quadrant yielding -3/2+1/2=-1/2, not 0.

`x=(2pi)/3` is a solution since `sqrt(3)sin(2pi)/3+cos(2pi)/3=sqrt(3)sqrt(3)/2+1/2=3/2+1/2=1`

So the solutions are given by `x=0+-2pi,(2pi)/3+-2pi`

or `x=0+-360^circ,x=120^circ+-360^circ` .