cứu em với ạ

Giải thích các bước giải:

a.Ta có:

$y'=(\sin(2x+3))' =2\cos(2x+3)$

$\to 2\cos(2x+3)=0$

$\to \cos(2x+3)=0$

$\to 2x+3=\dfrac12\pi+k\pi, k\in Z$

$\to x=\dfrac12(\dfrac12\pi+k\pi-3)$

b.Ta có:

$y'=(2\sin x+3)'=2\cos(x)$

$\to 2\cos(x)=0$

$\to \cos(x)=0$

$\to x=\dfrac12\pi+k\pi, k\in Z$

c.Ta có:

$y'=((3-2x)^4)'=4\left(3-2x\right)^3\left(3-2x\right)'\:=-8\left(3-2x\right)^3$

$\to -8(3-2x)^3=0$

$\to 3-2x=0$

$\to x=\dfrac32$

d.Ta có:

$y'=(\cos(4x+5))'=-\sin \left(4x+5\right)\left(4x+5\right)'\:=-\sin \left(4x+5\right)\cdot \:4$

$\to -\sin \left(4x+5\right)\cdot \:4=0$

$\to \sin(4x+5)=0$

$\to 4x+5=k\pi, k\in Z$

$\to 4x=k\pi-5$

$\to x=\dfrac14(k\pi-5)$