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- Feb 14, 2012

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\(\displaystyle \sqrt{3}\sin x(\cos x-\sin x)+(2-\sqrt{6})\cos x+2\sin x+\sqrt{3}-2\sqrt{2}=0\)

- Thread starter anemone
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- Feb 14, 2012

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\(\displaystyle \sqrt{3}\sin x(\cos x-\sin x)+(2-\sqrt{6})\cos x+2\sin x+\sqrt{3}-2\sqrt{2}=0\)

- Aug 30, 2012

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I'm beginning to think that you are a sadist...

\(\displaystyle \sqrt{3}\sin x(\cos x-\sin x)+(2-\sqrt{6})\cos x+2\sin x+\sqrt{3}-2\sqrt{2}=0\)

-Dan

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\(\displaystyle \sqrt{3}\sin(x)(\cos(x)-\sin(x))+(2-\sqrt{6})\cos(x)+2\sin(x)+\sqrt{3}-2\sqrt{2}=0\)

First, let's do the distribution dance...

\(\displaystyle \sqrt{3}\sin(x)\cos(x)-\sqrt{3}\sin^2(x)+2\cos(x)-\sqrt{6}\cos(x)+2\sin(x)+\sqrt{3}-2\sqrt{2}=0\)

Now, regroup...

\(\displaystyle \sqrt{3}(\sin(x)\cos(x)+1-\sin^2(x))+2(\sin(x)+\cos(x))-\sqrt{6}\cos(x)-2\sqrt{2}=0\)

Within the second factor of the first term, apply a Pythagorean identity:

\(\displaystyle \sqrt{3}(\sin(x)\cos(x)+\cos^2(x))+2(\sin(x)+\cos(x)-\sqrt{2}-\sqrt{6}\cos(x)=0\)

Now factor:

\(\displaystyle \sqrt{3}\cos(x)(\sin(x)+\cos(x))+2(\sin(x)+\cos(x)-\sqrt{2}(\sqrt{3}\cos(x)+2)=0\)

\(\displaystyle (\sin(x)+\cos(x))(\sqrt{3}\cos(x)+2)-\sqrt{2}(\sqrt{3}\cos(x)+2)=0\)

\(\displaystyle (\sin(x)+\cos(x)-\sqrt{2})(\sqrt{3}\cos(x)+2)=0\)

The second factor has no real root, hence we are left with:

\(\displaystyle \sin(x)+\cos(x)=\sqrt{2}\)

\(\displaystyle \sin\left(x+\frac{\pi}{4} \right)=1\)

\(\displaystyle x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

\(\displaystyle x=\frac{\pi}{4}+2k\pi=\frac{\pi}{4}(8k+1)\)

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- Feb 14, 2012

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Thank you

\(\displaystyle \sqrt{3}\sin(x)(\cos(x)-\sin(x))+(2-\sqrt{6})\cos(x)+2\sin(x)+\sqrt{3}-2\sqrt{2}=0\)

First, let's do the distribution dance...

\(\displaystyle \sqrt{3}\sin(x)\cos(x)-\sqrt{3}\sin^2(x)+2\cos(x)-\sqrt{6}\cos(x)+2\sin(x)+\sqrt{3}-2\sqrt{2}=0\)

Now, regroup...

\(\displaystyle \sqrt{3}(\sin(x)\cos(x)+1-\sin^2(x))+2(\sin(x)+\cos(x))-\sqrt{6}\cos(x)-2\sqrt{2}=0\)

Within the second factor of the first term, apply a Pythagorean identity:

\(\displaystyle \sqrt{3}(\sin(x)\cos(x)+\cos^2(x))+2(\sin(x)+\cos(x)-\sqrt{2}-\sqrt{6}\cos(x)=0\)

Now factor:

\(\displaystyle \sqrt{3}\cos(x)(\sin(x)+\cos(x))+2(\sin(x)+\cos(x)-\sqrt{2}(\sqrt{3}\cos(x)+2)=0\)

\(\displaystyle (\sin(x)+\cos(x))(\sqrt{3}\cos(x)+2)-\sqrt{2}(\sqrt{3}\cos(x)+2)=0\)

\(\displaystyle (\sin(x)+\cos(x)-\sqrt{2})(\sqrt{3}\cos(x)+2)=0\)

The second factor has no real root, hence we are left with:

\(\displaystyle \sin(x)+\cos(x)=\sqrt{2}\)

\(\displaystyle \sin\left(x+\frac{\pi}{4} \right)=1\)

\(\displaystyle x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi\) where \(\displaystyle k\in\mathbb{Z}\)

\(\displaystyle x=\frac{\pi}{4}+2k\pi=\frac{\pi}{4}(8k+1)\)

My solution is quite similar to yours but I figure it is only fair if I posted mine as well here...

My solution:

We're given

\(\displaystyle \sqrt{3}\sin(x)(\cos(x)-\sin(x))+(2-\sqrt{6})\cos(x)+2\sin(x)+\sqrt{3}-2\sqrt{2}=0\)

I noticed there is another term \(\displaystyle (2\sin x)\) which I could group it to the first term, after I multiplied \(\displaystyle \sqrt{3}\) onto everything inside the parentheses...

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)-\sqrt{3}\sin(x)+2)+(\sqrt{2}\sqrt{2}-\sqrt{2}\sqrt{3})\cos(x)+\sqrt{3}-2\sqrt{2}=0\)

Also, the second term can be simplified further to get

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)-\sqrt{3}\sin(x)+2)+\sqrt{2}\cos(x)(\sqrt{2}-\sqrt{3})+\sqrt{3}-2\sqrt{2}=0\)

And we can rewrite\(\displaystyle \sqrt{3}\) as \(\displaystyle (\sqrt{3})(1)=\sqrt{3}\left(\sin^2(x)+\cos^2 (x) \right)=\sqrt{3}\sin^2(x)+\sqrt{3}\cos^2(x)\) and observe that each of these two terms can be simplified out by collecting them to the first and second terms of the equation above:

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)-\sqrt{3}\sin(x)+2)+\sqrt{2}\cos(x)(\sqrt{2}-\sqrt{3})+\sqrt{3}\sin^2(x)+\sqrt{3}\cos^2(x)-2\sqrt{2}=0\)

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)-\sqrt{3}\sin(x)+2+\sqrt{3}\sin(x))+\sqrt{2}\cos(x)\left(\sqrt{2}-\sqrt{3}+\frac{\sqrt{3}}{\sqrt{2}}\cos^2(x) \right)-2\sqrt{2}=0\)

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)+2)+\sqrt{2}\cos(x) \left(\frac{2-\sqrt{2}\sqrt{3}+\sqrt{3}\cos(x)}{\sqrt{2}} \right)-2\sqrt{2}=0\)

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)+2)+\cos(x)(2-\sqrt{2}\sqrt{3}+\sqrt{3}\cos(x))-2\sqrt{2}=0\)

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)+2)+\cos(x)(2-\sqrt{2}\sqrt{3}+\sqrt{3}\cos(x))-2\sqrt{2}=0\)

\(\displaystyle \sin(x)(\sqrt{3}\cos(x)+2)+\cos(x)(\sqrt{3}\cos(x)+2)-\sqrt{2}\sqrt{3}\cos(x)-2\sqrt{2}=0\)

\(\displaystyle (\sqrt{3}\cos(x)+2)(\sin(x)+\cos(x))-\sqrt{2}(\sqrt{3}\cos(x)+2)=0\)

\(\displaystyle (\sqrt{3}\cos(x)+2)(\sin(x)+\cos(x)-\sqrt{2})=0\)

At this point, I think we can just refer to your post on how to solve for the values for \(\displaystyle x\).

Well, I'm usually pretty hard onI'm beginning to think that you are a sadist...

-Dan

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