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

1 Answer
Narad T.
Nov 8, 2016

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

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