6tan^2x-4sin^2x-1=6sin^2x/cos^2x-4sin^2x-1设t=cos^2x,则sin^2x=1-t,所以上式销缺饥变为6(1-t)/t-4(1-t)-1=[6(1-t)-4t(1-t)-t]/t=(4t^2-11t+6)/t=[(t-2)(4t-3)]/t=0时t=2或t=3/亏返4,经检验t=2舍扮巧去,所以t=3/4,即cos^2x=3/4=(1+cos2x)/2,即cos2x=1/2所以2x=2kπ±π/3(k∈Z),即x=kπ±π/6(k∈Z)即为所求