anonymous
  • anonymous
2 sin u cos u - 2√2 sin u - cos u +√2=0. Find all solutions in the interval {0,360°) any ideas where to start?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
start by rewriting \[2sin(u)cos(u) =sin(2u)\]?
anonymous
  • anonymous
yikes. this one is a real pain. but it is doable i think. first replace sin(u) by x, cos(u) by y to rewrite as \[2xy-2\sqrt{2}x-y+\sqrt{2}=0\] \[2xy+\sqrt{2}=2\sqrt{2}x+y\] now square both sides. \[(2xy+\sqrt{2})^2=(2\sqrt{2}x+y)^2\] \[4x^2y^2+2\sqrt{2}xy+2=8x^2+4\sqrt{2}xy+y^2\] \[4x^2y^2+2=8x^2+y^2\]
anonymous
  • anonymous
now since \[cos^2(u)=1-sin^2(u)\] replace \[y^2\] by \[1-x^2\] in the above equation. \[4x^2(1-x^2)+2=8x^2+(1-x^2)\] \[4x^2-4x^4=7x^2+1\] which is a quadratic equation in \[x^2\] so set = 0 and solve \[4x^4-3x^2-1=0\] \[(4x^2+1)(x^2-1)=0\]

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anonymous
  • anonymous
oops i made a mistake. it is \[4x^4+3x^2-1=0\] \[(4x^2-1)(x^2+1)=0\] \[x^2+1=0\]has not solutions \[4x^2-1=0\] \[4x^2=1\] \[x^2=\frac{1}{4}\] \[x=\pm\frac{1}{2}\]
anonymous
  • anonymous
\[sin(x)=\frac{1}{2}\] \[x=30\] or \[x=150\]
anonymous
  • anonymous
\[sin(u)=-\frac{1}{2}\] \[u=210\] \[u=330\]
anonymous
  • anonymous
and now, (believe it or not we are not done) you have to check the answers because one step was to square both sides, so it is possible that one of the solutions was introduced during the squaring. i will let you do that.
anonymous
  • anonymous
this was so long it makes me think that there is a snappier way to do it. if there is let me know.
anonymous
  • anonymous
i made a typo right in the beginning but it does not change the answer. in the second line \[(2xy+\sqrt{2})^2=4x^2y^2+4\sqrt{2}xy+2\] thats is why you can subtract the \[4\sqrt{2}xy\] from both sides.

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