Cho f(x)=sinx+1/3sin3x+2/5sin5x. Giải phương trình: f′(x)=0

Hướng dẫn giải:

Ta có:  \( {f}'(x)=\cos x+\cos 3x+2\cos 5x \)

 \( {f}'(x)=0\Leftrightarrow \cos x+\cos 3x+2\cos 5x=0\Leftrightarrow (\cos x+\cos 5x)+(\cos 3x+\cos 5x)=0 \)

 \( \Leftrightarrow 2\cos 3x\cos 2x+2\cos 4x\cos x=0\Leftrightarrow (4{{\cos }^{3}}x-3\cos x)\cos 2x+(2{{\cos }^{2}}2x-1)\cos x=0 \)

 \( \Leftrightarrow \left[ (4{{\cos }^{2}}x-3)\cos 2x+2{{\cos }^{2}}2x-1 \right]\cos x=0 \)

\(\Leftrightarrow \left[ \left( 2(1+\cos 2x)-3 \right)\cos 2x+2{{\cos }^{2}}2x-1 \right]\cos x=0\)

\(\Leftrightarrow \left[ \begin{align}  & \cos x=0 \\  & \left( 2(1+\cos 2x)-3 \right)\cos 2x+2{{\cos }^{2}}2x-1=0 \\ \end{align} \right.\) \( \Leftrightarrow \left[ \begin{align}  & \cos x=0 \\  & 4{{\cos }^{2}}2x-\cos 2x-1=0 \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & \cos x=0 \\  & \cos 2x=\frac{1\pm \sqrt{17}}{8} \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{2}+k\pi  \\  & 2x=\pm \arccos \left( \frac{1+\sqrt{17}}{8} \right)+k2\pi  \\  & 2x=\pm \arccos \left( \frac{1-\sqrt{17}}{8} \right)+k2\pi  \\ \end{align} \right. \)

 \( \Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{2}+k\pi  \\  & x=\pm \frac{1}{2}\arccos \left( \frac{1+\sqrt{17}}{8} \right)+k\pi  \\  & x=\pm \frac{1}{2}\arccos \left( \frac{1-\sqrt{17}}{8} \right)+k\pi  \\ \end{align} \right.,\text{ }k\in \mathbb{Z} \).