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

Question

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

1 Answer

Impact of this question

3979 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-08T19:45:18

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#

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question