Question #9162b

1 Answer
Narad T.
Nov 25, 2017

The solutions are #S={7/6pi,3/2pi,11/6pi}#

Explanation:

The equation is

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

Factorising this quadratic equation

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

Therefore,

#2sinx-1=0# and #sinx+1=0#

#sinx=1/2# and #sinx=-1#

#sinx=1/2#, #=>#, #x=7/6pi# and #x=11/6pi# , #[mod2pi]#

#sinx=-1#, #=>#, #x=3/2pi#, #[mod 2pi]#

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