2sin^(2)x-sinx=1 Find the general form and all solutions in the interval [0,2pi) ?

1 Answer
maganbhai P.
Feb 27, 2018

#x=kpi+(-1)^kpi/2,kinZ# or
#x=kpi-(-1)^kpi/6,kinZ#

Explanation:

#2sin^2x-sinx=1#
#2sin^2x-2sinx+sinx-1=0#
2sinx(sinx-1)+1(sinx-1)=0
(sinx-1)(2sinx+1)=0
#sinx=1 or sinx=-1/2#
#sinx=sin(pi/2) or sinx=sin(-pi/6)#
So,general solution :
#x=kpi+(-1)^kpi/2,kinZ# or
#x=kpi-(-1)^kpi/6,kinZ#
If ,# x in[0,2pi]# then
#sinx=1rArrx=pi/2,#
#sinx=(-1/2)=sin(-pi/6)rArrx=(7pi)/6,(11pi)/6#
where, #pi/2,(7pi)/6,(11pi)/6in[0.2pi]#

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