Which values are two of the possible solutions to the equation? sine (5 x minus pi) = startfraction startroot 2 endroot over 2 endfraction, 0 less-than-or-equal-to x less-than 2 pi left-brace startfraction pi over 20 endfraction, startfraction 3 pi over 20 endfraction right-brace left-brace startfraction pi over 4 endfraction, startfraction 7 pi over 4 endfraction right-brace left-brace startfraction pi over 4 endfraction, startfraction 7 pi over 20 endfraction right-brace left-brace startfraction 5 pi over 4 endfraction, startfraction 7 pi over 4 endfraction right-brace.