anonymous
  • anonymous
solve 3 sin^2 x - 2 = cos x
Mathematics
katieb
  • katieb
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tyteen4a03
  • tyteen4a03
Remember that \(sin^2(x) + cos^2(x) = 1\).
anonymous
  • anonymous
yes...
anonymous
  • anonymous
Is 3 sin^2(x-2)=cos x

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tyteen4a03
  • tyteen4a03
@ashwinjohn3 I believe it's 3sin^2(x) - 2.
anonymous
  • anonymous
yea its 3 sin^2 (x) - 2 = cos (x)
hartnn
  • hartnn
change sin^2 x to cos^2 x using \(\sin^2(x) + \cos^2(x) = 1\) then if you put cos^2 x = y, you'll get a quadratic in y.
anonymous
  • anonymous
ok... then,here \[\sin ^{2} x=1-\cos ^{2} x\] \[3 (1-\cos ^{2}x )-2=3-3 \cos ^{2}x -2=1-3\cos ^{2}x\]
anonymous
  • anonymous
thats what i get but then i cant figure out how to solve the angle :/
hartnn
  • hartnn
put cos x =y, then you get a quadratic in y..
anonymous
  • anonymous
But \[1=\cos ^{2}x+\sin ^{2} x\] =\[\sin ^{2}x+\cos ^{2}x-3\cos ^{2}x=\sin ^{2}x-2\cos ^{2}x\]
hartnn
  • hartnn
?
hartnn
  • hartnn
\(3 \sin^2 x - 2 = \cos x \\ 1-3\cos^2x = \cos x \\ 1-3y^2 =y. \\ 3y^2 +y-1=0 \) can you solve this quadratic ?
anonymous
  • anonymous
yeaaa but by using calculator, the answer is in decimal point. is there any other way i can calculate the angle?
hartnn
  • hartnn
you'll get same answer using any other method, that you got on calculator. whatever you got, equate it to y= ..., ... so, cos x = ... , .... then take cos^{-1} (or arccos) of those 2 values ("decimal values"), using calculator, and ou'll get 2 angles.
hartnn
  • hartnn
you might be getting, -0.767 and 0.434 then 2 angles will be cos^{-1}(-0.767) and cos^{-1}(0.434) calculate those using calculator.
anonymous
  • anonymous
hmmm okay thanks
hartnn
  • hartnn
welcome :)

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