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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
`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`
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`
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