anonymous
  • anonymous
Find all solutions to the equation sin^2(x)cos^2(x) = 1/4.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
cos(2x) - sin(x) = 1 cos²(x) - sin²(x) - sin(x) = 1 1 - cos²(x) + sin²(x) + sin(x) = 0 2sin²(x) + sin(x) = 0 sin(x)(2sin(x) + 1) = 0 sin(x) = 0 x = nπ OR sin(x) = -1/2 x = (4n-1)π/2 ± π/3 Between 0 and 2π: 0, π, 7π/6, 11π/6, 2π
.Sam.
  • .Sam.
sin^2(x)cos^2(x) = 1/4 (1-cos^2(x))cos^2(x) = 1/4 -cos^4(x)+cos^2(x)-1/4=0 cos^(2)x=1/2 cosx=sqrt(2)/2 x=45,315
lgbasallote
  • lgbasallote
hey @.Sam. i am learning with the asker here and i don't get -cos^4(x)+cos^2(x)-1/4=0 cos^(2)x=1/2 cosx=sqrt(2)/2 pls explain for me? :)

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anonymous
  • anonymous
sin^2(x)cos^2(x) = 1/4. (1/4)(2sin cos)^2 =1/4 2sin cos = 1 or -1 sin2x =1 or -1 Could it be done like this?
anonymous
  • anonymous
ya it can be its near 2 the ans
.Sam.
  • .Sam.
-cos^4(x)+cos^2(x)-1/4=0 Factor, let u=cos x -u^(2)+u-(1)/(4)=0 u^(2)-u+(1)/(4)=0 u=(-b+-sqrt(b^(2)-4ac))/(2a) where au^(2)+bu+c=0 u=(-(-1)+-sqrt((-1)^(2)-4(1)((1)/(4))))/(2(1)) u=(1+-sqrt((-1)^(2)-4(1)((1)/(4))))/(2(1)) u=1/2 cos^(2)x=1/2
lgbasallote
  • lgbasallote
hey @Coco..how did you have 1/4 in the left hand side? your transitions are fast :p sorry...just want to learn along with the asker here hehe..i think he doesnt get either
lgbasallote
  • lgbasallote
oh ok! sam :D thanks
anonymous
  • anonymous
sin 2x = 2sinx cos x we have sin^2(x)cos^2(x) =(sinx cos x)^2 in order to express it in sin2x, (2 sinx cosx /2)^2 =(1/4) (sin2x)^2

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