Question

1. Sin³x+cos³x=0

2. Sin(x-π/3)+2cos(x-π/6)=0

Answer 1

1: \(\Leftrightarrow\sin^3x=-\cos^3x\)

\(\Leftrightarrow\sin^3x=-\sin^3\left(\dfrac{\Pi}{2}-x\right)\)

\(\Leftrightarrow\sin^3x=\sin^3\left(-\dfrac{\Pi}{2}+x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\Pi}{2}+x+k2\Pi\\x=\dfrac{\Pi}{2}-x+k2\Pi\end{matrix}\right.\Leftrightarrow x=\dfrac{\Pi}{4}+k\Pi\)

2: \(\Leftrightarrow-\dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x=0\)

\(\Leftrightarrow\sin x\cdot\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\cdot\cos x=0\)

\(\Leftrightarrow\sin x\cdot\dfrac{\cos\Pi}{6}-\cos x\cdot\sin\left(\dfrac{\Pi}{6}\right)=0\)

\(\Leftrightarrow\sin\left(x-\dfrac{\Pi}{6}\right)=0\)

\(\Leftrightarrow x-\dfrac{\Pi}{6}=k\Pi\)

hay \(x=k\Pi+\dfrac{\Pi}{6}\)