What are the general solution of equation #2sin^2x-sinx-1=0#?

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Answer 1 · 2016-08-01T12:53:37

Roots are given by #x=npi-(-1)^n(pi/6}#

and #x=2npi+pi/2#, where #n# is an integer

Roots of #2sin^2x-sinx-1=0# can be obtained by factorizing the function.

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

#hArr2sin^2x-2sinx+sinx-1=0#

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

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

and roots of #2sin^2x-sinx-1=0# are given by

Hence, either #2sinx+1=0# i.e. #sinx=-1/2# and

#x=npi-(-1)^n(pi/6}#

or #sinx-1=0# i.e. #sinx=1# and

#x=2npi+pi/2#, where #n# is an integer