Since #sin^2x+cos^2x = 1# , #sin^6x le sin^2 x# and #cos^10xle cos^2 x# (the last two following from the fact that both #sin^2x# and #cos^2x# are bounded above by 1, we have
# sin^6x+cos^10x le 1 #
and thus
#2x^2 le sin^6x+cos^10x-1 le 0# is only possible if