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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.
 10 months ago
 10 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.
 10 months ago
 10 months ago

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

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

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@bagajr ??
 10 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
 10 months ago

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

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

kc_kennylau Group TitleBest ResponseYou've already chosen the best response.0
Corollary: \(\sin^2x=1\cos^2x\) and \(\cos^2x=1\sin^2x\).
 10 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.
 10 months ago

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

cupcakezz Group TitleBest ResponseYou've already chosen the best response.0
@bagajr
 10 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 \(1\sin^2x\).
 10 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
 10 months ago

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

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

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

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

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

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

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

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

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

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

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