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- Feb 14, 2012

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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.