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Question

If the equation sin4x(k+2)sin2x(k+3)=0 has a solution then k must lie in the interval:

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Solution

sin4x(k+2)sin2x(k+3)=0
sin2x[0,1]
put sin2x=t,t[0,1]
t2(k+2)t(k+3)=0
put t=0
0(k+2)0k3=0
k=3
put t=1
1k2k3=0
2k=4
k=2
k[3,2]

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