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

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Algebraic!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)
 one year ago

dragonslyBest ResponseYou've already chosen the best response.0
the intervals are between 0 and 2pie
 one year ago

dragonslyBest ResponseYou've already chosen the best response.0
and wouldn't it be root .49 because sin is squrd?
 one year ago

Algebraic!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)
 one year ago

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