MHB How Can You Solve Complex Trigonometric Equations?

[anemone]

Solve the equation

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

 

[topsquark]

Solve the equation

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

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

-Dan

 

[MarkFL]

Here is my solution:

Spoiler

We are given to solve:

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

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

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

Now, regroup...

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

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

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

Now factor:

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

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

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

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

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

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

$$x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi$$ where $$k\in\mathbb{Z}$$

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

 

[anemone]

Here is my solution:

Spoiler

We are given to solve:

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

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

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

Now, regroup...

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

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

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

Now factor:

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

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

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

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

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

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

$$x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi$$ where $$k\in\mathbb{Z}$$

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

Thank you MarkFL for participating and your result is correct!:D

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

My solution:

Spoiler

We're given

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

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

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

Also, the second term can be simplified further to get

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

And we can rewrite$$\sqrt{3}$$ as $$(\sqrt{3})(1)=\sqrt{3}\left(\sin^2(x)+\cos^2 (x) \right)=\sqrt{3}\sin^2(x)+\sqrt{3}\cos^2(x)$$ 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:

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

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

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

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

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

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

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

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

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

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

-Dan

Well, I'm usually pretty hard on myself...(Tongueout)