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

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Algebraic! Group TitleBest 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)
 2 years ago

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

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

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
good call...
 2 years ago

Algebraic! Group TitleBest 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)
 2 years ago

dragonsly Group TitleBest 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)
 2 years ago
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