Tìm x∈[0;14] nghiệm đúng phương trình: cos3x−4cos2x+3cosx−4=0

Hướng dẫn giải:

Ta có:  \( (*)\Leftrightarrow \left( 4{{\cos }^{3}}x-3cosx \right)-4(2{{\cos }^{2}}x-1)+3\cos x-4=0 \)

 \( \Leftrightarrow 4{{\cos }^{3}}x-8cos{{x}^{2}}=0\Leftrightarrow 4{{\cos }^{2}}x(cosx-2)=0 \)

 \( \Leftrightarrow \left[ \begin{align}  & \cos x=0\text{ }(n) \\  & \cos x=2\text{ }(\ell ) \\ \end{align} \right. \) (Do  \( -1\le \cos x\le 1 \))

 \( \Leftrightarrow x=\frac{\pi }{2}+k\pi \text{ }(k\in \mathbb{Z}) \)

Ta có:  \( x\in [0;14]\Leftrightarrow 0\le \frac{\pi }{2}+k\pi \le 14\Leftrightarrow -\frac{\pi }{2}\le k\pi \le 14-\frac{\pi }{2}\Leftrightarrow -\frac{1}{2}\le k\le \frac{14}{\pi }-\frac{1}{2}\approx 3,9 \)

Mà  \( k\in \mathbb{Z} \) nên  \( k\in \{0;1;2;3\} \). Do đó: \(x\in \left\{ \frac{\pi }{2};\frac{3\pi }{2};\frac{5\pi }{2};\frac{7\pi }{2} \right\}\).