MHB Solve sin³(x) - cos³(x) = sin²(x)

[anemone]

Hi MHB,

This seemingly very simple and straightforward trigonometric equation (Solve for x for which sin³x-cos³x=sin²x) has me stumped! I am fretting after I exhausted all kind of tricks that I could think of (which I will mention in my attempts next) trying to attempt the problem and now, I wanted so much to ask for help here at MHB!

First attempt:

I first divided through the equation by sin³x, bearing in mind sin³x≠0 and obtained:

$$1-\cot^3x=cscx$$

$$(1-\cot^3x)^2=\csc^2x$$

$$(1-\cot^3x)^2=1+\cot^2x$$

If l let $$ \cot x=k,$$ the equation above becomes $$(1-k^3)^2=1+k^2$$ and this is a futile attempt to say the least.

Second attempt:

I applied the triple angle formulas for both sine and cosine to the given equation and ended up with

$$\frac{3\sin x-\sin 3x}{4}-\frac{\cos3x+3\cos x}{4}=\sin^2x$$

$$3\sin x-\sin 3x-\cos 3x-3\cos x=2(2\sin^2x)$$

$$3(\sin x-\cos x)-(\sin3x+\cos3x)=2(1-\cos2x)
$$
This is another unwise substitution to make, knowing it can hardly bring us any closer to the answer...

Anyway, this is how I approached it.

$$3( \sin x- \cos x)-(\sin3x+\cos3x)=2(1-\cos2x)$$

$$3\sqrt{2}\sin(x-\frac{\pi}{4}-\sqrt{2}\sin(3x+\frac{\pi}{4})=2(1-\cos 2x)$$ I stopped right at this point, because we can tell this is a very bad move...

Third attempt:

$$\sin^3 x-\cos^3 x=\sin^2 x$$

$$\sin^2 x(\sin x-1)=\cos x\cos^2 x=\cos x(1-\sin^2 x)=\cos x(1+\sin x)(1-\sin x)$$

$$\sin^2 x(\sin x-1)-\cos x(1+\sin x)(1-\sin x)=0$$

$$\sin^2 x(\sin x-1)+\cos x(1+\sin x)(\sin x-1)=0$$

$$(\sin^2 x+\cos x(1+\sin x))(\sin x-1)=0$$

$$(\sin^2 x+\cos x+\sin x\cos x))(\sin x-1)=0$$

and obviously $$\sin x=1$$ is an answer to the problem.

And now, all kind of thoughts enter my mind, I am not even sure when we have the case ab=0, if a=0 is imminent and true, then b isn't necessary a zero. And everything seems to reach an impasse after getting sinx=1 as one of the answers to the problem.

So, I decided to factor the LHS of the given equation and see how far I can go with this approach...

$$(\sin x-\cos x)(\sin^2 x+\sin x\cos x+\cos^2 x)=\sin^2 x$$

$$(\sin x-\cos x)(1+\sin x\cos x)=\sin^2 x$$

and it's also safe to say at this juncture this won't work out well and I am at my wit's end to solve this problem.

I would appreciate it if someone could offer me some hints to solve it.

Thanks.

 

[issacnewton]

Re: Solve sin³x-cos³x=sin²x

Hello
This is how I did it. We have,
\[ \sin^3x - \cos^3 x = \sin^2 x \]
Write this as the following.
\[ \sin^3x-\sqrt{(1-\sin^2x)}\;(1-\sin^2x) = \sin^2x \]
Here let \( t = \sin x\), and then rearranging the terms we get a polynomial in t.
\( (t^3 - t^2)^2 = (1-t^2)^3 \). Expanding this we have,
\[ 2t^6 - 2t^5 - 2t^4 + 3t^2 - 1 = 0 \]
Now using the rational root theorem and synthetic division, we can arrive at \( (t-1)^2 \) as
one factor. So obviously \(t=1\) is one root. The remainder polynomial equation then is
\[ 2t^4+2t^3 - 2t -1 = 0 \]
This is a quartic equation, for which there are known solutions. For this equation there
are two real and two imaginary solutions. The real solutions are \(t = 0.903408 \) and
\(t = -0.582522 \). Plugging back in the original equation for \( \sin x\), we can solve for \(x\).
\( x = 1.12765 \) and \( x = 3.76342 \), also the first solution of \( t=1 \) leads to \( x = \frac{\pi}{2}\).
So three solutions are \(x = 1.12765\;\;x = 3.76342\;\;x = \frac{\pi}{2} \)

 

[Dustinsfl]

Re: Solve sin³x-cos³x=sin²x

Hello
This is how I did it. We have,
\[ \sin^3x - \cos^3 x = \sin^2 x \]
Write this as the following.
\[ \sin^3x-\sqrt{(1-\sin^2x)}\;(1-\sin^2x) = \sin^2x \]
Here let \( t = \sin x\), and then rearranging the terms we get a polynomial in t.
\( (t^3 - t^2)^2 = (1-t^2)^3 \). Expanding this we have,
\[ 2t^6 - 2t^5 - 2t^4 + 3t^2 - 1 = 0 \]
Now using the rational root theorem and synthetic division, we can arrive at \( (t-1)^2 \) as
one factor. So obviously \(t=1\) is one root. The remainder polynomial equation then is
\[ 2t^4+2t^3 - 2t -1 = 0 \]
This is a quartic equation, for which there are known solutions. For this equation there
are two real and two imaginary solutions. The real solutions are \(t = 0.903408 \) and
\(t = -0.582522 \). Plugging back in the original equation for \( \sin x\), we can solve for \(x\).
\( x = 1.12765 \) and \( x = 3.76342 \), also the first solution of \( t=1 \) leads to \( x = \frac{\pi}{2}\).
So three solutions are \(x = 1.12765\;\;x = 3.76342\;\;x = \frac{\pi}{2} \)

I think you need to add \(+ 2\pi k\) where \(k\in\mathbb{Z}\) since there are infinitely many solutions for each principle argument.

 

[issacnewton]

Re: Solve sin³x-cos³x=sin²x

I am assuming that \( x \in [0,2\pi] \)

 

[Opalg]

Re: Solve sin³x-cos³x=sin²x

Hello
This is how I did it. We have,
\[ \sin^3x - \cos^3 x = \sin^2 x \]
Write this as the following.
\[ \sin^3x-\sqrt{(1-\sin^2x)}\;(1-\sin^2x) = \sin^2x \]
Here let \( t = \sin x\), and then rearranging the terms we get a polynomial in t.
\( (t^3 - t^2)^2 = (1-t^2)^3 \). Expanding this we have,
\[ 2t^6 - 2t^5 - 2t^4 + 3t^2 - 1 = 0 \]
Now using the rational root theorem and synthetic division, we can arrive at \( (t-1)^2 \) as
one factor. So obviously \(t=1\) is one root. The remainder polynomial equation then is
\[ 2t^4+2t^3 - 2t -1 = 0 \]
This is a quartic equation, for which there are known solutions. For this equation there
are two real and two imaginary solutions. The real solutions are \(t = 0.903408 \) and
\(t = -0.582522 \). Plugging back in the original equation for \( \sin x\), we can solve for \(x\).
\( x = 1.12765 \) and \( x = 3.76342 \), also the first solution of \( t=1 \) leads to \( x = \frac{\pi}{2}\).
So three solutions are \(x = 1.12765\;\;x = 3.76342\;\;x = \frac{\pi}{2} \)

I agree with the solutions $x = 3.76342$ and $x = \frac{\pi}{2}$, but not with $x = 1.12765$. The reason is that when going from $\sin x$ to $x$ you need to ensure that you choose the angle in the correct quadrant. When you check against the original equation $\sin^3x - \cos^3x = \sin^2x$, you find that both solutions $\sin x = 0.903408$ and $\sin x = -0.582522$ require $\cos x$ to be negative. So in both cases you need to have $x$ in the interval $\bigl[\frac\pi2,\frac{3\pi}2\bigr]$. Therefore the solution is not $x = 1.12765$ but $x = \pi - 1.12765 = 2.01394$.

Ideally, it would be far better to have an exact solution rather than these numerical approximations. But I am not convinced that that can be done for this problem.

 

[issacnewton]

Re: Solve sin³x-cos³x=sin²x

Opalg, thanks for pointing mistake. I did the math in hurry.
Since you are from UK, you must have heard of S.L. Loney who did write a book
on trigonometry. I think this problem could be from that book.