\[ \begin{eqnarray} \sin^3 x + \sin x \cos x + \cos^3 x& = & 1 \\
\sin^3 x + \sin x \cos x & = & 1- \cos^3 x \\
\sin x (\sin^2 x + \cos x )& = & 1- \cos^3 x \\
\sin x & = & \frac {1- \cos^3 x}{\sin^2 x + \cos x } \\
\sin x & = & \frac {1- \cos^3 x}{1 -\cos^2 x+ \cos x } \\ \end{eqnarray} \]

Square both sides, substitute \( \sin^2 x = 1 - \cos^2 x \), expand and simplify and substitute \(y = \cos x \), you we get an equation \( y(y-1)(y^4 - y^2 + 1) = 0 \). Can you take it from here?

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TopNewest\[ \begin{eqnarray} \sin^3 x + \sin x \cos x + \cos^3 x& = & 1 \\ \sin^3 x + \sin x \cos x & = & 1- \cos^3 x \\ \sin x (\sin^2 x + \cos x )& = & 1- \cos^3 x \\ \sin x & = & \frac {1- \cos^3 x}{\sin^2 x + \cos x } \\ \sin x & = & \frac {1- \cos^3 x}{1 -\cos^2 x+ \cos x } \\ \end{eqnarray} \]

Square both sides, substitute \( \sin^2 x = 1 - \cos^2 x \), expand and simplify and substitute \(y = \cos x \), you we get an equation \( y(y-1)(y^4 - y^2 + 1) = 0 \). Can you take it from here?

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Simplify it using \(a^3+b^3\) formula and using identities to \(\sin x +\cos x=1\). I think you know how to take it from here.

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Also answer in my other notes.

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