Question #ee035

1 Answer
Jan 25, 2018

#{2kpi+pi/2}uu{2kpi-7pi/6}, k in ZZ#.

Explanation:

I do not know about the "added corner" method, but, hope

my following solution may help!

#sqrt3sinx-2cosx=sqrt3-cosx#.

#rArr sqrt3sinx-2cosx+cosx=sqrt3#.

#rArr sqrt3sinx-cosx=sqrt3, or, cosx-sqrt3sinx=-sqrt3#.

Multiplying by #1/2,# we get,

#1/2cosx-sqrt3/2sinx=-sqrt3/2, i.e., #

#cosxcos(pi/3)-sinxsin(pi/3)=cos(pi-pi/6)#.

#:. cos(x+pi/3)=cos(5pi/6).........<<1>>#.

Recall that, #costheta=cosalpha rArr theta=2kpi+-alpha, k in ZZ#.

Hence, from #<<1>>,# we have,

#x+pi/3=2kpi+-5pi/6, k in ZZ.#

#:. x=2kpi+-5pi/6-pi/3, k in ZZ#.

Thus, #x=2kpi+5pi/6-pi/3=2kpi+pi/2, or, #

#x=2kpi-5pi/6-pi/3=2kpi-7pi/6, kin ZZ#.

So, the Solution Set is, #{2kpi+pi/2}uu{2kpi-7pi/6}, k in ZZ#.