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

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

1 Answer
Jan 6, 2018

AA

Explanation:

sin^2(x)-2sin(x)+k=0sin2(x)2sin(x)+k=0

Let u=sin(x)u=sin(x)

Then:

u^2-2u+k=0u22u+k=0

Using the discriminant of a quadratic:

b^2-4acb24ac

IF:

b^2-4ac>0b24ac>0 ( real and different roots )

b^2-4ac=0b24ac=0 ( real and repeated roots )

b^2-4ac<0b24ac<0 ( complex roots )

For no real solution we need:

b^2-4ac<0b24ac<0

Plugging in values:

(-2)^2-4(1)(k)<0(2)24(1)(k)<0

4-4k<044k<0

k>1k>1

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

i.e. color(white)(88)1< k< oo881<k<