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cupcakezz
 one year ago
2cos^2x + sinx1=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.
cupcakezz
 one year ago
2cos^2x + sinx1=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.

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RaphaelFilgueiras
 one year ago
Best ResponseYou've already chosen the best response.0dw:1389329041541:dw

cupcakezz
 one year ago
Best ResponseYou've already chosen the best response.0can you send some things so I can start looking at it ?? @bagajr

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.0Firstly, be familiarized with this identity: \(\sin^2x+\cos^2x=1\)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.0Corollary: \(\sin^2x=1\cos^2x\) and \(\cos^2x=1\sin^2x\).

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.0You may want to convert the question into an ordinary quadratic equation.

cupcakezz
 one year ago
Best ResponseYou've already chosen the best response.0But why cant it be factored??

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.0You will know why when you convert the question into an ordinary quadratic equation by changing \(\cos^2x\) to \(1\sin^2x\).

bagajr
 one year ago
Best ResponseYou've already chosen the best response.2a) 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.\[2\cos^{2}(x)+\sin(x)1=0\]\[2[1\sin^{2}(x)]+\sin(x)1=0\]\[22\sin^{2}(x)+\sin(x)1=0\]\[2\sin^{2}(x)+\sin(x)+1=0\]\[2\sin^{2}(x)\sin(x)1=0\]c) This isn't necessary, but you can make factoring easier by setting y=sin(x)\[2y^{2}y1=0\]\[(2y+1)(y1)=0\]\[(2\sin(x)+1)(\sin(x)1)=0\]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

cupcakezz
 one year ago
Best ResponseYou've already chosen the best response.0@bagajr wow thank you very much !

bagajr
 one year ago
Best ResponseYou've already chosen the best response.2No problem, I love helping people with this stuff.

cupcakezz
 one year ago
Best ResponseYou've already chosen the best response.0@bagajr could you help me with one more?

bagajr
 one year ago
Best ResponseYou've already chosen the best response.2Is it a question about this problem or a different one?

cupcakezz
 one year ago
Best ResponseYou've already chosen the best response.0prove: tan^2x(1+1/tan^2x)=1/(1sin^2x)

cupcakezz
 one year ago
Best ResponseYou've already chosen the best response.0@bagajr what is the answer??

bagajr
 one year ago
Best ResponseYou've already chosen the best response.2\[\tan^2(x)[1+\frac{1}{\tan^{2}(x)}]=\frac{1}{1\sin^{2}(x)}\]\[\tan^{2}(x)+1=\frac{1}{\cos^{2}(x)}\]\[\frac{\sin^{2}(x)}{\cos^{2}(x)}+1=\frac{1}{\cos^{2}(x)}\]\[\sin^{2}(x)+\cos^{2}(x)=1\]\[1=1\]

bagajr
 one year ago
Best ResponseYou've already chosen the best response.2Sorry it takes so long and you're welcome

cupcakezz
 one year ago
Best ResponseYou've already chosen the best response.0its fine, no worries
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