# TrigonometrySolve sin³(x) - cos³(x) = sin²(x)

#### anemone

##### MHB POTW Director
Staff member
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:

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

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

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

If l let $$\displaystyle \cot x=k,$$ the equation above becomes $$\displaystyle (1-k^3)^2=1+k^2$$ 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

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

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

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

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

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

Third attempt:

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

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

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

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

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

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

and obviously $$\displaystyle \sin x=1$$ 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...

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

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

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

##### Member
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 $$x = 1.12765\;\;x = 3.76342\;\;x = \frac{\pi}{2}$$

#### dwsmith

##### Well-known member
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

##### Member
Re: Solve sin³x-cos³x=sin²x

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

#### Opalg

##### MHB Oldtimer
Staff member
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

##### Member
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.