Solve for x: √3 sin x + cosx = 1
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`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` .
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