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cupcakezz
Group Title
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.
 7 months ago
 7 months ago
cupcakezz Group Title
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.
 7 months ago
 7 months ago

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cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@Luigi0210
 7 months ago

RaphaelFilgueiras Group TitleBest ResponseYou've already chosen the best response.0
dw:1389329041541:dw
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@bagajr ??
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
can you send some things so I can start looking at it ?? @bagajr
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@kc_kennylau
 7 months ago

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

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Corollary: \(\sin^2x=1\cos^2x\) and \(\cos^2x=1\sin^2x\).
 7 months ago

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
You may want to convert the question into an ordinary quadratic equation.
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
But why cant it be factored??
 7 months ago

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

bagajr Group TitleBest ResponseYou've already chosen the best response.2
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.\[2\cos^{2}(x)+\sin(x)1=0\]\[2[1\sin^{2}(x)]+\sin(x)1=0\]\[22\sin^{2}(x)+\sin(x)1=0\]\[2\sin^{2}(x)+\sin(x)+1=0\]\[2\sin^{2}(x)\sin(x)1=0\]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
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@bagajr wow thank you very much !
 7 months ago

bagajr Group TitleBest ResponseYou've already chosen the best response.2
No problem, I love helping people with this stuff.
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@bagajr could you help me with one more?
 7 months ago

bagajr Group TitleBest ResponseYou've already chosen the best response.2
Is it a question about this problem or a different one?
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
prove: tan^2x(1+1/tan^2x)=1/(1sin^2x)
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@bagajr what is the answer??
 7 months ago

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

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
thanks again !
 7 months ago

bagajr Group TitleBest ResponseYou've already chosen the best response.2
Sorry it takes so long and you're welcome
 7 months ago

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
its fine, no worries
 7 months ago
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