a. Với mọi \(x \in\mathbb R\), ta có:
\(y' = 2\cos 2x + 2\sin x = 2\left( {1 - 2{{\sin }^2}x} \right) + 2\sin x\)
\(=-4{{\sin }^2}x+2\sin x+2\)
Vậy \(y' = 0 \Leftrightarrow 2{\sin ^2}x - \sin x - 1 = 0\)
\( \Leftrightarrow \left[ {\matrix{ {\sin x = 1} \cr {\sin x = -{1 \over 2}} \cr } } \right. \Leftrightarrow \left[ {\matrix{ {x = {\pi \over 2} + k2\pi } \cr {x = - {\pi \over 6} + k2\pi } \cr {x = {{7\pi } \over 6} + k2\pi } \cr }\left( {k \in \mathbb Z} \right) } \right.\)
b. Với mọi \(x \in\mathbb R\), ta có: \(y' = 6\cos 2x - 8\sin 2x + 10\)
Vậy \(y' = 0 \Leftrightarrow 4\sin 2x - 3\cos 2x = 5\)
\( \Leftrightarrow {4 \over 5}\sin 2x - {3 \over 5}\cos 2x = 1\,\,\left( 1 \right)\)
Vì \({\left( {{4 \over 5}} \right)^2} + {\left( {{3 \over 5}} \right)^2} = 1\) nên có số \(α\) sao cho \(\cos \alpha = {4 \over 5}\,\text{ và }\,\sin \alpha = {3 \over 5}\)
Thay vào (1), ta được :
\(\eqalign{ & \sin 2x\cos \alpha - sin\alpha cos2x = 1 \cr & \Leftrightarrow \sin \left( {2x - \alpha } \right) = 1 \cr & \Leftrightarrow 2x - \alpha = {\pi \over 2} + k2\pi \cr & \Leftrightarrow x = {1 \over 2}\left( {\alpha + {\pi \over 2} + k2\pi } \right)\,\,\left( {k \in\mathbb Z} \right) \cr} \)
c. Với mọi \(x \in\mathbb R\), ta có: \(y' = - 2\cos x{\mathop{\rm sinx}\nolimits} + cosx = cosx\left( {1 - 2\sin x} \right)\)
\(\eqalign{ & y' = 0 \Leftrightarrow \cos x\left( {1 - 2\sin x} \right) = 0 \Leftrightarrow \left[ {\matrix{ { \cos x = 0 } \cr {1 - 2\sin x = 0 } \cr } } \right. \cr & \Leftrightarrow \left[ {\matrix{ {x = {\pi \over 2} + k\pi} \cr {{\mathop{\rm sinx}\nolimits} = {1 \over 2} \Leftrightarrow \left[ {\matrix{ {x = {\pi \over 6} + k2\pi } \cr {x = {{5\pi } \over 6} + k2\pi } \cr } } \right. } \cr } } \right. \cr} \)
Vậy \(x = {\pi \over 2} + k\pi ;x = {\pi \over 6} + k2\pi ;x = {{5\pi } \over 6} + k2\pi \left( {k \in\mathbb Z} \right)\)
d.
\(\eqalign{ & y' = {1 \over {{{\cos }^2}x}} - {1 \over {{{\sin }^2}x}}\,\forall\,x \ne k{\pi \over 2} \cr & y' = 0 \Leftrightarrow {1 \over {{{\cos }^2}x}} = {1 \over {{{\sin }^2}x}} \Leftrightarrow {\tan ^2}x = 1 \cr & \Leftrightarrow \tan x = \pm 1 \Leftrightarrow x = \pm {\pi \over 4} + k\pi ,k \in \mathbb Z \cr} \)
