anonymous
  • anonymous
2cos^2x + sinx-1=0 a. Explain why this equation cannot be factored. b. Use a trigonometric identity to change the equation into one that can be factored. c. Factor the equation. d. Determine all solutions in the interval 0≤x≤2π.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
|dw:1389329041541:dw|
anonymous
  • anonymous

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anonymous
  • anonymous
can you send some things so I can start looking at it ?? @bagajr
anonymous
  • anonymous
kc_kennylau
  • kc_kennylau
Firstly, be familiarized with this identity: \(\sin^2x+\cos^2x=1\)
kc_kennylau
  • kc_kennylau
Corollary: \(\sin^2x=1-\cos^2x\) and \(\cos^2x=1-\sin^2x\).
kc_kennylau
  • kc_kennylau
You may want to convert the question into an ordinary quadratic equation.
anonymous
  • anonymous
But why cant it be factored??
anonymous
  • anonymous
kc_kennylau
  • kc_kennylau
You will know why when you convert the question into an ordinary quadratic equation by changing \(\cos^2x\) to \(1-\sin^2x\).
anonymous
  • anonymous
a) Because cos^2x and sinx are not the same variable. b) Use the following identity to solve for cos^2x\[\sin^{2}(x)+\cos^{2}(x)=1\]\[\cos^{2}(x) =1-\sin^{2}(x)\]Substitute cos^2x for this new expression.\[2\cos^{2}(x)+\sin(x)-1=0\]\[2[1-\sin^{2}(x)]+\sin(x)-1=0\]\[2-2\sin^{2}(x)+\sin(x)-1=0\]\[-2\sin^{2}(x)+\sin(x)+1=0\]\[2\sin^{2}(x)-\sin(x)-1=0\]c) This isn't necessary, but you can make factoring easier by setting y=sin(x)\[2y^{2}-y-1=0\]\[(2y+1)(y-1)=0\]\[(2\sin(x)+1)(\sin(x)-1)=0\]d) When you multiply two things to get 0, one of those things must be 0.\[2\sin(x)+1=0\]\[\sin(x)-1=0\]Solve both equations to get that sin(x) is -1/2 or 1. Use unit circle to find solutions, which are pi/2, 5pi/4, and 7pi/4. |dw:1389329795828:dw|
anonymous
  • anonymous
@bagajr wow thank you very much !
anonymous
  • anonymous
No problem, I love helping people with this stuff.
anonymous
  • anonymous
@bagajr could you help me with one more?
anonymous
  • anonymous
Is it a question about this problem or a different one?
anonymous
  • anonymous
prove: tan^2x(1+1/tan^2x)=1/(1-sin^2x)
anonymous
  • anonymous
anonymous
  • anonymous
@bagajr what is the answer??
anonymous
  • anonymous
\[\tan^2(x)[1+\frac{1}{\tan^{2}(x)}]=\frac{1}{1-\sin^{2}(x)}\]\[\tan^{2}(x)+1=\frac{1}{\cos^{2}(x)}\]\[\frac{\sin^{2}(x)}{\cos^{2}(x)}+1=\frac{1}{\cos^{2}(x)}\]\[\sin^{2}(x)+\cos^{2}(x)=1\]\[1=1\]
anonymous
  • anonymous
thanks again !
anonymous
  • anonymous
Sorry it takes so long and you're welcome
anonymous
  • anonymous
its fine, no worries

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