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# Help me!!

The general solution of this equation is

Note by Ashley Shamidha
2 years, 1 month ago

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

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? · 2 years, 1 month ago

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. · 2 years, 1 month ago