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anonymous
 5 years ago
sin 2x= 1/2
anonymous
 5 years ago
sin 2x= 1/2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because 1/2=sin30 = sin 2x therefore 2x=30 x=15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also, sine is positive in the second quadrant, so sine of 2x=(18030) = 1/2 also. So x=75 degrees is a solution too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0These are principle solutions. The general solution is \[\theta = n.180^o+(1)^n30^o\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats wrong i need it in the interval [0, 2 pi)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can convert from degrees to radians. It's equivalent, not wrong. \[180^o = \pi^c \rightarrow 1^o=\frac{\pi^c}{180}\rightarrow 15^o=\frac{15\pi}{180}=\frac{\pi}{12}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would 5 pi / 12 be equl to it to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[75^o=\frac{75\pi}{180}=\frac{5\pi}{12}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I need the multiple angle equation of x/2 = negative square root of 3/ 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos x/2=  square root 3 /2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cosine in negative in the second and third quadrants. So, given cos(theta)=sqrt(3)/2 implies theta = 30 degrees, the angles needed here are: (i) 18030 = 150 (ii) 180+30 = 210

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So (i) x = 300 (ii) x = 420 which you convert to radians, and then subtract integral multiples of 2pi until the angle lies in your principal domain [0,2pi) here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah i need it in the principal domain [0, 2 pi)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0please hurry up this test is due in 15 minutes and i am clueless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey gut that has been helping me

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wtf, i'm not your lackey. just put down 5pi/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, calm down loki ^_^
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