Q.
If x∈[0,2π],a,b,c∈R(t>0) such that
1+sin2x+∣∣t+t4−4∣∣+(a+b+c)2≤0, then the value of
∣∣sinxta3+b3+c3−cosx23abctsinxcosxa2+b2+c2∣∣
Solution:
1+sin2x=0,t+t4−4=0,a+b+c=0
⇒sinx+cosx=0,t=2
use C1→C1−C2, we get
∣∣ccc000−cosx23abctsinxcosxa2+b2+c2∣∣=0