(1)f'(x)=6cosx(sinx)^5-6sinx(cosx)^5+4kcosx(sinx)^3-4ksinx(cosx)^3
=2sinxcosx[(sinx)^2-(cosx)^2](3+2k)=-(3+2k)sin2xcos2x
當(dāng)3+2k=0,即k=-3/2時(shí),f'(x)=0
(2)f(x)=[(sinx)^2+(cosx)^2][(sinx)^4-(sinxcosx)^2+(cosx)^4]+k((sin x)^4 + (cos x)^4)=(1+k)(sinx)^4-(sinxcosx)^2+(1+k)(cosx)^4
令t=(sinx)^2∈[0,1],則(cosx)^2=1-t∈[0,1],所以
f(t)=(1+k)t^2+t(t-1)+(1+k)(t-1)^2=(3+2k)t^2-(3+2k)t+(1+k)
=8t^2/5-8t/5+3/10=0,所以2t-1=±1/2,所以t=1/4或t=3/4
所以sinx=1/2或根號(hào)3/2[負(fù)值已舍],所以a=30°(π/6)或a=60°(π/3)
(3)f(x)=0有解的條件是判別式△≥0,所以(3+2k)^2-4(3+2k)(1+k)≥0
化為(2k+1)(2k+3)≤0,所以k的取值范圍是[-3/2,-1/2]