a) Ta có: \(f(x) = \sin^3 2x\)
\(⇒ f’(x) = 3\sin^2 2x (\sin2x)’ = 6\sin^2 2x \cos2x\)
Do đó:
\(\eqalign{
& f'(x) = g(x)\cr& \Leftrightarrow 6si{n^2}2x\cos 2x = 4\cos 2x - 5\sin 4x \cr
& \Leftrightarrow 6si{n^2}2x\cos 2x = 4\cos 2x - 10\sin 2x\cos 2x \cr
& \Leftrightarrow \cos 2x(3{\sin ^2}2x + 5\sin 2x - 2) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\cos 2x = 0\,\,\,\,\,\,(1) \hfill \cr
3{\sin ^2}2x + 5\sin 2x - 2 = 0 \,\,\,\, (2)\hfill \cr} \right. \cr} \)
Giải (1): \(2x = {\pi \over 2} + k\pi \,\,(k \in \mathbb Z) \Leftrightarrow x = {\pi \over 4} + {{k\pi } \over 2} (k \in \mathbb Z)\)
Giải (2): \( \Leftrightarrow \left[ \begin{array}{l}\sin 2x = - 2\,\,\left( {ktm} \right)\\\sin 2x = \frac{1}{3}\,\,\,\,\,\left( {tm} \right)\end{array} \right.\)
\(\eqalign{
& \sin 2x = {1 \over 3} \Leftrightarrow \left[ \matrix{
2x = \arcsin ({1 \over 3}) + k2\pi \hfill \cr
2x = \pi - \arcsin ({1 \over 3}) + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {1 \over 2}\arcsin ({1 \over 3}) + {k\pi } \hfill \cr
x = {\pi \over 2} - {1 \over 2}\arcsin ({1 \over 2}) + {k\pi } \hfill \cr} \right.;k \in \mathbb Z \cr} \)
Tóm lại, phương trình đã cho có ba nghiệm là:
\(\left[ \matrix{
x = {\pi \over 4} + {{k\pi } \over 2} \hfill \cr
x = {1 \over 2}\arcsin ({1 \over 3}) + {k\pi } \hfill \cr
x = {\pi \over 2} - {1 \over 2}\arcsin ({1 \over 2}) + {k\pi } \hfill \cr} \right.;k \in \mathbb Z\)
b) Ta có: \(f’(x) = -60sin 3x – 60 sin 5x + 60 sin4x = 0\)
Do đó:
\(\eqalign{
& f'(x) = 0 \Leftrightarrow - \sin 3x - \sin 5x + \sin 4x = 0 \cr
& \Leftrightarrow \sin 5x + \sin 3x - \sin 4x=0 \cr
& \Leftrightarrow 2\sin 4x{\mathop{\rm cosx}\nolimits} - sin4x = 0 \cr
& \Leftrightarrow sin4x(2cosx - 1) = 0 \cr} \)
\(\eqalign{
& \Leftrightarrow \left[ \matrix{
\sin 4x = 0 \hfill \cr
{\mathop{\rm cosx}\nolimits} = {1 \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
4x = k\pi \hfill \cr
x = \pm {\pi \over 3} + k2\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = k{\pi \over 4} \hfill \cr
x = \pm {\pi \over 3} + k2\pi \hfill \cr} \right.;k \in\mathbb Z \cr}\)
