General solution for 2sin^2x+cosx=12sin2x+cosx=1?

1 Answer
Jun 5, 2018

x={2kpi+-(2pi)/3,kinZZ}uu{2kpi,kinZZ}

Explanation:

Here,

2sin^2x+cosx=1

=>2(1-cos^2x)+cosx=1

=>2-2cos^2x+cosx=1

=>2cos^2x-cosx-1=0

=>2cos^2x-2cosx+cosx-1=0

=>2cosx(cosx-1)+1(cosx-1)=0

=>(2cosx+1)(cosx-1)=0

=>2cosx=-1 or cosx=1

=>cosx=-1/2 or cosx=1

(i)cosx=-1/2=cos(pi-pi/3)

=>cosx=cos((2pi)/3)

=>x=2kpi+-(2pi)/3,kinZZ

(ii)cosx=1=>x=2kpi,kinZZ

Hence, the general solution is :

x={2kpi+-(2pi)/3,kinZZ}uu{2kpi,kinZZ}