#2sin^2x-sinx-1=0#?

1 Answer
maganbhai P.
Apr 25, 2018

#x={(4k+1)pi/2,kinZZ}uu {kpi-(-1)^kpi/6,kinZZ}#

Explanation:

This is your first question and it is incomplete.

I tried to solve the trig. equn.

#2sin^2x-sinx-1=0#

#=>2sin^2x-2sinx+sinx-1=0#

#=>2sinx(sinx-1)+1(sinx-1)=0#

#=>(sinx-1)(2sinx+1)=0#

#=>sinx-1=0 or 2sinx+1=0#

#=>sinx=1 or sinx=-1/2#

Now,

#(i)sinx=1=>color(blue)(x=(4k+1)pi/2,kinZZ#

#(ii)sinx=-1/2=>sinx=sin(-pi/6)#

#:.x=kpi+(-1)^k(-pi/6),kinZZ#

#=>color(blue)(x=kpi-(-1)^kpi/6,kinZZ#

Hence,

#x={(4k+1)pi/2,kinZZ}uu {kpi-(-1)^kpi/6,kinZZ}#

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