Solve the equation:Sin(x) + root3*cos(x) = 1

Solve the equation: `sin x + sqrt(3)cos x=1`
.

`sinx+sqrt(3)cosx=1` Given
equation

`sin^2x+2sqrt(3)sinxcosx+3cos^2x=1` Square both
sides

`sin^2x+cos^2x+2sqrt(3)sinxcosx+2cos^2x=1`

`1+2sqrt(3)sinxcosx+2cos^2x=1`
Pythagorean identity

`2sqrt(3)sinxcosx=-2cos^2x` Subtract
from both sides

`sqrt(3)sinx=-cosx` Divide by `2cosx`
;`cosx!=0`

`tanx=-1/sqrt(3)`

Here
`x=(5pi)/6 +- kpi` with `k in ZZ` . However, squaring both sides introduced an
extraneous solution. `(5pi)/6` does not work.

Thus the
solution is `x=(11pi)/6 +- 2kpi` ;`k in ZZ`