a)
\(\eqalign{
& 2\sin {x \over 2}{\cos ^2}x - 2\sin {x \over 2}{\sin ^2}x = {\cos ^2}x - {\sin ^2}x \cr
& \Leftrightarrow 2\sin {x \over 2}({\cos ^2}x - {\sin ^2}x) = {\cos ^2}x - {\sin ^2}x \cr
& \Leftrightarrow 2\sin {x \over 2}.cos2x = \cos 2x\cr& \Leftrightarrow \cos 2x(2\sin {x \over 2} - 1) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos 2x = 0 \hfill \cr
\sin {x \over 2} = {1 \over 2} = \sin {\pi \over 6} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
2x = {\pi \over 2} + k\pi \hfill \cr
\left[ \matrix{
{x \over 2} = {\pi \over 6} + k2\pi \hfill \cr
{x \over 2} = \pi - {\pi \over 6} + k2\pi \hfill \cr} \right. \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over 4} + \frac{k\pi}{2} \hfill \cr
x = {\pi \over 3} + k4\pi \hfill \cr
x = {{5\pi } \over 3} + k4\pi \hfill \cr} \right.(k \in\mathbb Z) \cr} \)
b) Ta có:
\(\eqalign{
& 3cos{\rm{ }}x + 4sin{\rm{ }}x = 5 \cr
& \Leftrightarrow {3 \over 5}\cos x + {4 \over 5}\sin x = 1 \cr
& \Leftrightarrow \cos x\cos \varphi + \sin x\sin \varphi = 1\cr&(\text { với }cos\varphi = {3 \over 5};\sin \varphi = {4 \over 5}) \cr
& \Leftrightarrow \cos (x - \varphi ) = 1 \cr
& \Leftrightarrow x - \varphi = k2\pi \,\,\,(k \in\mathbb Z) \cr
& \Leftrightarrow x = \varphi + k2\pi \,\,\,(k \in\mathbb Z)\cr} \)
\(c) \,\,sin x + cosx = 1 + sinx. cosx\)
\(⇔ sin x – sin x. cosx + cosx – 1= 0\)
\(⇔ sin x ( 1 – cosx) – (1 – cosx) = 0\)
\(\eqalign{
& \Leftrightarrow (1 - \cos x)(\sin x - 1) = 0 \cr
& \Leftrightarrow \left[ \matrix{
{\mathop{\rm cosx}\nolimits} = 1 \hfill \cr
sinx = 1 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = k2\pi \hfill \cr
x = {\pi \over 2} + k2\pi \hfill \cr} \right.(k \in \mathbb Z) \cr} \)
d) Điều kiện \(\sin x ≥ 0\). Khi đó:
\(\eqalign{
& \sqrt {1 - \cos x} = \sin x \cr
& \Leftrightarrow 1-\cos x = {\sin ^2}x \cr
& \Leftrightarrow 1 - {\sin ^2}x - \cos x = 0 \cr
& \Leftrightarrow {\cos ^2}x - \cos x = 0 \cr
& \Leftrightarrow \cos x(cosx - 1) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos x = 0 \hfill \cr
\cos x = 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\pi \over 2} + k\pi \hfill \cr
x = k2\pi \hfill \cr} \right.;k \in\mathbb Z \cr}\)
\(\begin{array}{l}\pi \le \frac{\pi }{2} + k\pi \le 3\pi \\ \Leftrightarrow \frac{1}{2} \le k \le \frac{5}{2} \\ \mathop \Rightarrow \limits^{k \in Z} \left[ \begin{array}{l}k = 1 \Rightarrow x = \frac{{3\pi }}{2}\,\,\left( {ktm\,\,\sin x \ge 0} \right)\\k = 2\,\,\left( {tm} \right)\end{array} \right.\\\pi \le k2\pi \le 3\pi \\ \Leftrightarrow \frac{1}{2} \le k \le \frac{3}{2}\mathop \Rightarrow \limits^{k \in Z} k = 1 \Rightarrow x = 2\pi \,\,\left( {tm} \right)\end{array}\)
Vì \(\sin \frac{{5x}}{4} \le 1;\,\,\cos x \le 1 \Rightarrow \sin \frac{{5x}}{4} + \cos x \le 2 < 3 \Rightarrow \) phương trình trên vô nghiệm.
