# Equation with trigonometry

To solve $\sin^3x+\cos^3x=1$

So I just thought of a solution like:

Let $\sin x=t$. Then we have: $$t^3+(1-t^2) \sqrt{1-t^2} =1 \\ (1-t^2) \sqrt{1-t^2} =1-t^3 \\ (1-t^2)^3=(1-t^3)^2 \\ (1-t)^3(1+t)^3=(1-t)^2(1+t+t^2)^2 \\ (1-t)^2\left[ (1-t)(1+t)^3-(1+t+t^2)^2 \right]=0$$

Which is followed by $\sin x=0$ or $\sin x=1$. However, this solution seems very easy to make a silly mistake in (with all the squares and cubes of differences and sums). Is there any easier solution to this?

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Note that $\sin x$ and $\cos x$ are always at most $1$, so $\sin^3 x+\cos^3 x\leq \sin^2 x+\cos^2 x=1$ with equality only $\sin x,\cos x$ are $0$ or $1$.

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Hmm.. how do we get to $sin x = 0$ or $sin x = 1$ just from $sin^3x+cos^3x\leq1$? – Straightfw Apr 30 '12 at 18:05
Hint: If $\sin x<1$ and $\sin x\neq 0$ then $\sin^3 x < \sin^2 x$ @Straightfw – Thomas Andrews Apr 30 '12 at 18:09
Right! OHMIGOSH. Thank you very much, that sollution is brilliant :D – Straightfw Apr 30 '12 at 18:19