How do you solve #2( \sin ( t ) ) ^ { 2} - \sin ( t ) - 1= 0#?

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Answer 1 · 2017-01-01T11:23:18

The answer is #S={pi/2+2kpi,7/6pi+2kpi,16/6pi+2kpi}#, #(k in ZZ)#

We solve this as a quadratic equation

#ax^2+bx+c=0#

#2sin^2t-sint-1=0#

We calculate the discriminant

#Delta=b^2-4ac=(-1)^2-4(2)(-1)=1+8=9#

#Delta >0#, we have 2 real solutions

#x=(-b+-sqrtDelta)/(2a)#

#sint=(1+-3)/4#

#sint = 1#, #=>#, #t=pi/2+2kpi# #(kin ZZ)#

#sint=-1/2#, #=>#, #7/6pi+2kpi# and #=13/6+2kpi#, #(k in ZZ)#