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
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Luigi0210
anonymous
  • anonymous
|dw:1389329041541:dw|
anonymous
  • anonymous
@bagajr ??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
can you send some things so I can start looking at it ?? @bagajr
anonymous
  • anonymous
@kc_kennylau
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
@bagajr
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
@bagajr
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.