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Bài 3. Giải các phương trình
a) \(2\sin {x \over 2}{\cos ^2}x – 2\sin {x \over 2}{\sin ^2}x = {\cos ^2}x – {\sin ^2}x\)
b) \(3cos x + 4sin x = 5\)
c) \(sin x + cos x = 1 + sin x. cosx\)
d) \(\sqrt {1 – \cos x} = \sin x(x \in \left[ {\pi ,3\pi } \right]\)
e) \((cos{x \over 4} – 3\sin x)sinx + (1 + sin{x \over 4} – 3\cos x)cosx = 0\)
a)
\(\eqalign{
& 2\sin {x \over 2}{\cos ^2}x – 2\sin {x \over 2}{\sin ^2}x = {\cos ^2}x – {\sin ^2}x \cr
& \Leftrightarrow 2\sin {x \over 2}({\cos ^2}x – {\sin ^2}x) = {\cos ^2}x – {\sin ^2}x \cr
& \Leftrightarrow 2\sin {x \over 2}.cos2x = \cos 2x \Leftrightarrow \cos 2x(2\sin {x \over 2} – 1) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos 2x = 0 \hfill \cr
\sin {x \over 2} = {1 \over 2} = \sin {\pi \over 6} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
2x = {\pi \over 2} + k\pi \hfill \cr
\left[ \matrix{
{x \over 2} = {\pi \over 6} + k\pi \hfill \cr
{x \over 2} = \pi – {\pi \over 6} + k\pi \hfill \cr} \right. \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over 4} + k\pi \hfill \cr
x = {\pi \over 3} + k2\pi \hfill \cr
x = {{5\pi } \over 3} + k2\pi \hfill \cr} \right.(k \in\mathbb Z) \cr} \)
b) Ta có:
\(\eqalign{
& 3cos{\rm{ }}x + 4sin{\rm{ }}x = 5 \cr
& \Leftrightarrow {3 \over 5}\cos x + {4 \over 5}\sin x = 1 \cr
& \Leftrightarrow \cos x\cos \varphi + \sin x\sin \varphi = 1(\text { với }cos\varphi = {3 \over 5};\sin \varphi = {4 \over 5});(0 < \varphi < {\pi \over 2}) \cr
& \Leftrightarrow \cos (x – \varphi ) = 1 \cr
& \Leftrightarrow x – \varphi = k2\pi (k \in\mathbb Z) \cr
& \Leftrightarrow x = \varphi + k2\pi (k \in\mathbb Z)\cr} \)
c)
\(sin x + cosx = 1 + sinx. cosx\)
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\(⇔ sin x – sin x. cosx + cosx – 1= 0\)
\(⇔ sin x ( 1 – cosx) – (1 – cosx) = 0\)
\(\eqalign{
& \Leftrightarrow (1 – \cos x)(\sin x – 1) = 0 \cr
& \Leftrightarrow \left[ \matrix{
{\mathop{\rm cosx}\nolimits} = 1 \hfill \cr
sinx = 1 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = k2\pi \hfill \cr
x = {\pi \over 2} + k2\pi \hfill \cr} \right.(k \in \mathbb Z) \cr} \)
d) Điều kiện \(\sin x ≥ 0\). Khi đó:
\(\eqalign{
& \sqrt {1 – \cos x} = \sin x \cr
& \Leftrightarrow 1\cos x = {\sin ^2}x \cr
& \Leftrightarrow 1 – {\sin ^2}x – \cos x = 0 \cr
& \Leftrightarrow {\cos ^2}x – \cos x = 0 \cr
& \Leftrightarrow \cos x(cosx – 1) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos x = 0 \hfill \cr
\cos x = 1 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\pi \over 2} + k\pi \hfill \cr
x = k2\pi \hfill \cr} \right.;k \in\mathbb Z \cr}\)
Vì \( x ∈ [π, 3π]\) và \(sinx ≥ 0\) nên ta chọn:
\(\eqalign{
& k = 2 \Rightarrow x = {{5\pi } \over 2} \cr
& k = 1 \Rightarrow x = 2\pi \cr} \)
e)
\(\eqalign{
& (cos{x \over 4} – 3\sin x)sinx + (1 + sin{x \over 4} – 3\cos x)cosx = 0 \cr
& \Leftrightarrow \sin x\cos {x \over 4} + \cos x\sin {x \over 4} + \cos x – 3({\sin ^2}x + {\cos ^2}x) = 0 \cr
& \Leftrightarrow \sin (x + {x \over 4}) + \cos x – 3 = 0 \Leftrightarrow \sin {{5x} \over 4} + \cos x = 3 \cr} \)
Vì \(\sin {{5x} \over 4} \le 1\) , \(cosx ≤ 1\) nên phương trình trên vô nghiệm.