5. Giải các phương trình:

a) $2{\cos }^{2}x - 3\cos x +1 =0;$                    b) 25${\sin }^{2}x+15\sin 2x+9{\cos }^{2}x = 25$;

c) 2sinx + cosx = 1;                                  d) sinx + 1,5cotx = 0

Giải:

a) $2{\cos }^{2}x - 3\cos x +1=0\iff {[}_{\cos x=\frac{1}{2}}^{\cos x=1}\iff {[}_{x=\pm \frac{\mathrm{\pi }}{3}+k2\mathrm{\pi }}^{x=k2\mathrm{\pi }} (k\in \mathrm{\mathbb{Z}})$

$$\begin{gathered} b) 25{\sin }^{2}x+15\sin 2x+9{\cos }^{2}x=25 \\ \iff 25{\sin }^{2}x+15\sin 2x+9{\cos }^{2}x=25({\sin }^{2}x+{\cos }^{2}x) \\ \iff 16{\cos }^{2}x +30\sin x\cos x =0\iff 2\cos x(15\sin x-8\cos x)=0 \end{gathered}$$

$\iff {[}_{\tan x = \frac{8}{15}}^{\cos x = 0} {\iff }_{x= arc\tan \frac{8}{15}+k\mathrm{\pi }}^{x = \frac{\mathrm{\pi }}{2}+k\mathrm{\pi }} (k\in \mathrm{\mathbb{Z}})$

c) Chia hai vế phương trình cho $\sqrt{2^2+1^2}=\sqrt{5}$ ta được

$\frac{2}{\sqrt{5}}\sin x + \frac{1}{\sqrt{5}}\cos x=\frac{1}{\sqrt{5}}$

$Đặt\left\{\begin{array}{l}\cos \alpha =\frac{1}{\sqrt{5}}\\ \sin \alpha = \frac{2}{\sqrt{5}}\end{array}\right.ta có phương trình:$

$\cos x\cos \alpha +\sin x\sin \alpha =\cos \alpha \iff \cos (x-\alpha )=\cos \alpha$

$\iff {[}_{x-\alpha =-\alpha +k2\mathrm{\pi }}^{x-\alpha =\alpha +k2\mathrm{\pi }}\iff {[}_{x=k2\mathrm{\pi }}^{x=2\alpha +k2\mathrm{\pi }} (k\in \mathrm{\mathbb{Z}})$

d) Điều kiện : sinx ≠0 $\iff x\ne k\mathrm{\pi }$

sinx + 1,5cotx =0 $\iff \sin x +\frac{3}{2}.\frac{\cos x}{\sin x}=0$

$$\begin{gathered} \iff 2{\sin }^{2}x+3\cos x=0\iff 2(1-{\cos }^{2}x)+3\cos x=0 \\ \iff 2{\cos }^{2}x-3\cos x-2=0 \iff {[}_{\cos x = 2 \left(loại\right)}^{\cos x = -\frac{1}{2}} \end{gathered}$$

$\iff \cos x = \cos \frac{2\mathrm{\pi }}{3}\iff x=\pm \frac{2\mathrm{\pi }}{3}+k2\mathrm{\pi }, \mathrm{k}\in \mathrm{\mathbb{Z}}$