anonymous
  • anonymous
Find all solutions in the interval [0, 2π). 2 sin2x = sin x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Nnesha
  • Nnesha
set it equal to 0
Nnesha
  • Nnesha
let sin =x to make it easy :D \[\huge\rm 2 x^2 = x\]set it equal to 0 and then take out common factor
anonymous
  • anonymous
Can you elaborate more?

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Nnesha
  • Nnesha
|dw:1437488293143:dw| alright so first step ) subtract x b oth sides
Nnesha
  • Nnesha
let me know what yo uget
anonymous
  • anonymous
x^2 = 0
Nnesha
  • Nnesha
no
Nnesha
  • Nnesha
|dw:1437488395098:dw| -x stay same at left side
Nnesha
  • Nnesha
now factor that equation
anonymous
  • anonymous
X(2^2-1) = 0 X(4-1) = 0 X(3)=0
Nnesha
  • Nnesha
hmm it's x^2 not 2 to the 2 power
Nnesha
  • Nnesha
but yeah x is common factor so take it out \[\huge\rm x(2x-1)\] now set x and (2x-1) both equal to zero then solve for x
anonymous
  • anonymous
x= 1/2 so i would have to look at the unit circle to find the ones with 1/2 as the sine value between 0 and 2π or 0 and 180
Nnesha
  • Nnesha
not between 0 -180 0 -360
Nnesha
  • Nnesha
at what point sin x = 1/2 and sin x = 0 too
anonymous
  • anonymous
Yes. thank you
Nnesha
  • Nnesha
my pleasure :=)

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