Question

How do you solve `sqrt3cosx-sinx=1` in the interval [0,360]?

Answer 1

The answer is `(x=30º and x=330º)`

Explanation:

We transform the LHS into `cos(x+alpha)`
`cos(x+alpha)=cosxcosalpha-sinxsinalpha`
`rcos(x+alpha)=rcosxcosalpha-sinxsinalpha`
Comparing this to our equation
`rcosalpha=sqrt3` and `rsinalpha=1`
`r^2cos^2alpha+r^2sin^2alpha=3+1=4`
`r^2=4` `=>` `r=2`
`:. 2(sqrt3/2cosx-1/2sinx)=1`
`cosalpha=sqrt3/2` and `sinalpha=1/2`
`:. alpha=pi/6`
`cos(x+pi/6)=1/2`
`x+pi/6=pi/3`
`x=pi/6` and `x=11/6pi`