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How do you solve #2sin^2x-sinx-1=0 and find all exact general solutions?

1 Answer

Nov 24, 2016

#x=(7pi)/6+2kpi or x=(11pi)/6+2kpi or x=pi/2+2kpi#

Explanation:

By using the quadratic formula with respect to sin x, you get:

#sinx=(1+-sqrt(1-4*2*(-1)))/(2*2)#

#sinx=(1+sqrt(9))/4=(1+-3)/4#

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

that's

#x=(7pi)/6+2kpi or x=(11pi)/6+2kpi or x=pi/2+2kpi#

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