How to find the value of k give that equation has no solution?

Question

Can someone please explain to me how to do question 13? Thanks!

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Answer 1 · 2018-01-06T12:25:20

#A#

#sin^2(x)-2sin(x)+k=0#

Let #u=sin(x)#

Then:

#u^2-2u+k=0#

Using the discriminant of a quadratic:

#b^2-4ac#

IF:

#b^2-4ac>0# ( real and different roots )

#b^2-4ac=0# ( real and repeated roots )

#b^2-4ac<0# ( complex roots )

For no real solution we need:

#b^2-4ac<0#

Plugging in values:

#(-2)^2-4(1)(k)<0#

#4-4k<0#

#k>1#

#k in (1, oo)# ( open interval, endpoints not included )

i.e. #color(white)(88)1< k< oo#