anonymous
  • anonymous
Help with Trig-- sin X + 2 sin X cos X= 0 solve for all possible radian solutions (please explain each step) Thank you
Mathematics
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anonymous
  • anonymous
Help with Trig-- sin X + 2 sin X cos X= 0 solve for all possible radian solutions (please explain each step) Thank you
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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amistre64
  • amistre64
do you know your trig identities?
anonymous
  • anonymous
yes, 2 sin x cos x = sin 2A
amistre64
  • amistre64
good, then when does sin(x) + sin(2x) = 0? recall also that " -sin(x) = sin(-x)" that might be useful

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amistre64
  • amistre64
sin(2x) = -sin(x) sin(2x) = sin(-x) 2x = -x 2x + x = 0 3x = 0 .... when x = 0 thats what I come up with as a solution....
amistre64
  • amistre64
0 and pi i believe are the answers
anonymous
  • anonymous
thats what I got as an answer, but the answers in the back of the textbook says differently : pi k, 2 pi/3 +2 pi k, 4 pi/3 +2pi k
amistre64
  • amistre64
ahhh.... ALL possible solutions involves an interger constant of "k"
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amistre64
  • amistre64
there do appear to be 3 spots between 0 and 2pi that match up...
anonymous
  • anonymous
ohhhh
anonymous
  • anonymous
i see it, but i dont understand
anonymous
  • anonymous
how to get the points without knowing it
amistre64
  • amistre64
...yeah, trig is highly visual.
amistre64
  • amistre64
we could try this: sin(x) + 2 sin(x) cos(x)= 0 sin(x)(1 + 2cos(x)) = 0
amistre64
  • amistre64
sin(x) = 0 is a solution and (1+2cos(x)) = 0 are solutions
amistre64
  • amistre64
1+ 2 cos(x) = 0 2cos(x) = -1 cos(x) = -1/2
amistre64
  • amistre64
that works better :)
amistre64
  • amistre64
0, pi for sin(x) cos(x) is negative in Q2 and Q3...
amistre64
  • amistre64
cos(pi/3) = 1/2 .... 60 degrees. that would make it 120 degrees and 240 degrees as well.
amistre64
  • amistre64
2pi/3 + k2pi and 4pi/3 + k2pi...right?
anonymous
  • anonymous
yes, thank you very much!... I can see a light at the end of the tunnel!!! thank you
amistre64
  • amistre64
:) all it takes is a little persistents and alot of luck lol
anonymous
  • anonymous
thank you, your right about that!!

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