How Can You Solve Complex Trigonometric Equations?

[anemone]

Solve the equation

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

 

[topsquark]

Solve the equation

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

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

-Dan

 

[MarkFL]

Here is my solution:

Spoiler

We are given to solve:

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

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 \(\displaystyle k\in\mathbb{Z}\)

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

 

[anemone]

Here is my solution:

Spoiler

We are given to solve:

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

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 \(\displaystyle k\in\mathbb{Z}\)

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

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

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

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 \(\displaystyle x\).

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

-Dan

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

 

[CellCoree]

To solve this equation, we can first simplify the left side by factoring out a common factor of sin x:

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

Next, we can use the trigonometric identity \sin^2 x + \cos^2 x = 1 to simplify the expression inside the parentheses:

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

Now, we can use a substitution to make this equation a quadratic equation in terms of \cos x. Let's substitute u = \cos x, then the equation becomes:

\sqrt{3}u^3+2+\sqrt{3}-2\sqrt{2}=0

We can solve this quadratic equation by factoring or by using the quadratic formula. Either way, we get two solutions for u: u = \frac{-1+\sqrt{3}}{\sqrt{3}} and u = \frac{-1-\sqrt{3}}{\sqrt{3}}.

Since we have substituted u = \cos x, we can substitute back to find the solutions for x:

\cos x = \frac{-1+\sqrt{3}}{\sqrt{3}} and \cos x = \frac{-1-\sqrt{3}}{\sqrt{3}}

Using the inverse cosine function, we get the solutions:

x = \frac{\pi}{6} + 2n\pi and x = \frac{5\pi}{6} + 2n\pi, where n is any integer.

Therefore, the solutions to the original equation are:

x = \frac{\pi}{6} + 2n\pi, \frac{5\pi}{6} + 2n\pi, \frac{7\pi}{6} + 2n\pi, and \frac{11\