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cupcakezz
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.
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can you send some things so I can start looking at it ?? @bagajr
Firstly, be familiarized with this identity: \(\sin^2x+\cos^2x=1\)
Corollary: \(\sin^2x=1-\cos^2x\) and \(\cos^2x=1-\sin^2x\).
You may want to convert the question into an ordinary quadratic equation.
But why cant it be factored??
You will know why when you convert the question into an ordinary quadratic equation by changing \(\cos^2x\) to \(1-\sin^2x\).
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|
@bagajr wow thank you very much !
No problem, I love helping people with this stuff.
@bagajr could you help me with one more?
Is it a question about this problem or a different one?
prove: tan^2x(1+1/tan^2x)=1/(1-sin^2x)
@bagajr what is the answer??
\[\tan^2(x)[1+\frac{1}{\tan^{2}(x)}]=\frac{1}{1-\sin^{2}(x)}\]\[\tan^{2}(x)+1=\frac{1}{\cos^{2}(x)}\]\[\frac{\sin^{2}(x)}{\cos^{2}(x)}+1=\frac{1}{\cos^{2}(x)}\]\[\sin^{2}(x)+\cos^{2}(x)=1\]\[1=1\]
Sorry it takes so long and you're welcome