Phương trình cos 3x-cos 2x+9sinx-4=0 trên khoảng

* Hướng dẫn giải

Đáp án B

Ta có: $PT\iff 4c{\text{os}}^{3}x-3\cos x+2{\sin }^{2}x+9\sin x-5=0$ 

$\begin{array}{l}\iff \cos x\left(4c{\text{os}}^{2}x-3\right)+2{\sin }^{2}x+9\sin x-5=0\\ \iff \cos x\left(1-4{\sin }^{2}x\right)+\left(2\sin x-1\right)\left(\mathrm{s}\text{inx}+5\right)=0\iff \left(2\sin x-1\right)\left(\cos x+2\sin x\cos x+\mathrm{s}\text{inx}+5\right)=0\\ \iff \left(2\sin x-1\right)\left(\mathrm{s}\text{inx}+\cos x+\sin 2x+5\right)=0\iff 2\sin x-1=0\iff \mathrm{s}\text{inx}=\frac{1}{2}\iff \left[\begin{array}{l}x=\frac{\pi }{6}k2\pi \\ x=\frac{5\pi }{6}+k2\pi \end{array}\right.\end{array}$

Với $x\in \left(0;3\pi \right)\Rightarrow x=\left\{\frac{\pi }{6};\frac{5\pi }{6};\frac{\pi }{6}+2\pi ;\frac{5\pi }{6}+2\pi \right\}\Rightarrow T=6\pi$.