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anonymous
 4 years ago
Solve this trigonometric equation.
2sin^21=0
anonymous
 4 years ago
Solve this trigonometric equation. 2sin^21=0

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Note that this is the same as \[\cos (2x)=0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sin(x)=\pm\frac{\sqrt{2}}{2}\] is a start

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how didyou get cos(2x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Therefore \[x=\pi/4+n*\pi/2, n \in \mathbb{Z}\]

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0try 2sin^2x = 1 and then sin^2x = 1/2 then go from there

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0so\[\sin(x) = \sqrt{(1/2)} or \pm1/\sqrt{?}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but the restriction id 0 to pi/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327534375645:dw

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0you should know the angle that gives \[\sin(x)=1/\sqrt{2} or \sin ^{(1)} (1/\sqrt{2})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks i got it ur correct

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem..happy to help :):)
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