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

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Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.0(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)

dragonsly
 2 years ago
Best ResponseYou've already chosen the best response.0the intervals are between 0 and 2pie

dragonsly
 2 years ago
Best ResponseYou've already chosen the best response.0and wouldn't it be root .49 because sin is squrd?

Algebraic!
 2 years ago
Best ResponseYou've already chosen the best response.0(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)

dragonsly
 2 years ago
Best ResponseYou've already chosen the best response.0but how would you find the solution would i add 3.14 to root .49 and subtract 3.14 (btw everythings in radians)
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