how would you find the solution for this
sin^2x - .49=0

Question

how would you find the solution for this
sin^2x - .49=0<--- the answer has 4 different solutions

anonymous · Answer

(sinx)^2 =.49
sin(x) = +/- .49

doesn't give an interval, so it has infinite solutions...
if your interval is something like 0..2pi
then arcsin(-.49)   (two solutions)
and  arcsin(.49)     (two more solutions)

anonymous · Answer

the intervals are between 0 and 2pie

anonymous · Answer

and wouldn't it be root .49 because sin is squrd?

anonymous · Answer

good call...

anonymous · Answer

(sinx)^2 =.49
sin(x) = +/- sqrt(.49)

doesn't give an interval, so it has infinite solutions...
if your interval is something like 0..2pi
then arcsin(-sqrt(.49)   (two solutions)
and  arcsin(sqrt(.49) )    (two more solutions)

anonymous · Answer

but how would you find the solution would i add 3.14 to root .49 and subtract 3.14 (btw everythings in radians)