Giải các phương trình sau:

a) $\sin \left(\frac{x}{2}-\frac{\pi }{3}\right)=-\frac{\sqrt{3}}{4}$

b) $\sin \left(3x-30^{\circ }\right)=\sin 45^{\circ }$

c) $\sin \left(3x-\frac{3\pi }{4}\right)=\sin \left(\frac{\pi }{6}-x\right)$

d) $\sin \left(4x-\frac{\pi }{3}\right)=0$

e) $\cos \left(-x+\frac{\pi }{3}\right)=1$

f) $\cos \left(5x-\frac{\pi }{3}\right)=\sin \left(\frac{7\pi }{4}-2x\right)$

g) $\cos \left(2x+25^{\circ }\right)=-\frac{\sqrt{2}}{2}$

h) $\cos \left(\frac{\pi }{6}-2x\right)=-\frac{1}{4}$

(AI giải đề cho tôi)
Câu 128: Giải từng phương trình lượng giác cơ bản.

a) $\sin \left(\frac{x}{2}-\frac{\pi }{3}\right)=-\frac{\sqrt{3}}{4}$
$$\frac{x}{2}-\frac{\pi }{3}=(-1{)}^k\arcsin \left(-\frac{\sqrt{3}}{4}\right)+k\pi =(-1{)}^{k+1}\arcsin \frac{\sqrt{3}}{4}+k\pi$$
$$\Rightarrow x=\frac{2\pi }{3}+2(-1{)}^{k+1}\arcsin \frac{\sqrt{3}}{4}+2k\pi ,k\in \mathbb{Z}.$$

b) $\sin (3x-30^{\circ })=\sin 45^{\circ }$
$3x-30^{\circ }=45^{\circ }+k360^{\circ }$ hoặc $3x-30^{\circ }=135^{\circ }+k360^{\circ }$
$$\Rightarrow x=25^{\circ }+k120^{\circ }\text{hoặc}x=55^{\circ }+k120^{\circ },k\in \mathbb{Z}.$$

c) $\sin \left(3x-\frac{3\pi }{4}\right)=\sin \left(\frac{\pi }{6}-x\right)$
- Trường hợp 1: $3x-\frac{3\pi }{4}=\frac{\pi }{6}-x+2k\pi \Rightarrow 4x=\frac{11\pi }{12}+2k\pi \Rightarrow x=\frac{11\pi }{48}+\frac{k\pi }{2}$.
- Trường hợp 2: $3x-\frac{3\pi }{4}=\pi -\frac{\pi }{6}+x+2k\pi =\frac{5\pi }{6}+x+2k\pi \Rightarrow 2x=\frac{19\pi }{12}+2k\pi \Rightarrow x=\frac{19\pi }{24}+k\pi$.

Vậy $x=\frac{11\pi }{48}+\frac{k\pi }{2}$ hoặc $x=\frac{19\pi }{24}+k\pi$, $k\in \mathbb{Z}$.

d) $\sin \left(4x-\frac{\pi }{3}\right)=0\Rightarrow 4x-\frac{\pi }{3}=k\pi \Rightarrow x=\frac{\pi }{12}+\frac{k\pi }{4},k\in \mathbb{Z}.$

e) $\cos \left(-x+\frac{\pi }{3}\right)=1\Rightarrow -x+\frac{\pi }{3}=2k\pi \Rightarrow x=\frac{\pi }{3}-2k\pi ,k\in \mathbb{Z}.$

f) $\cos \left(5x-\frac{\pi }{3}\right)=\sin \left(\frac{7\pi }{4}-2x\right)=\cos \left(\frac{\pi }{2}-\frac{7\pi }{4}+2x\right)=\cos \left(2x-\frac{5\pi }{4}\right)$
- $5x-\frac{\pi }{3}=2x-\frac{5\pi }{4}+2k\pi \Rightarrow 3x=-\frac{11\pi }{12}+2k\pi \Rightarrow x=-\frac{11\pi }{36}+\frac{2k\pi }{3}$.
- $5x-\frac{\pi }{3}=-2x+\frac{5\pi }{4}+2k\pi \Rightarrow 7x=\frac{19\pi }{12}+2k\pi \Rightarrow x=\frac{19\pi }{84}+\frac{2k\pi }{7}$.

Vậy $x=-\frac{11\pi }{36}+\frac{2k\pi }{3}$ hoặc $x=\frac{19\pi }{84}+\frac{2k\pi }{7}$, $k\in \mathbb{Z}$.

g) $\cos (2x+25^{\circ })=-\frac{\sqrt{2}}{2}=\cos 135^{\circ }$
$2x+25^{\circ }=\pm 135^{\circ }+k360^{\circ }$
- $2x=110^{\circ }\Rightarrow x=55^{\circ }+k180^{\circ }$.
- $2x=-160^{\circ }\Rightarrow x=-80^{\circ }+k180^{\circ }$.

Vậy $x=55^{\circ }+k180^{\circ }$ hoặc $x=-80^{\circ }+k180^{\circ }$, $k\in \mathbb{Z}$.

h) $\cos \left(\frac{\pi }{6}-2x\right)=-\frac{1}{4}$
$$\frac{\pi }{6}-2x=\pm \arccos \left(-\frac{1}{4}\right)+2k\pi =\pm \left(\pi -\arccos \frac{1}{4}\right)+2k\pi$$
$$\Rightarrow x=\frac{\pi }{12}\mp \frac{1}{2}\left(\pi -\arccos \frac{1}{4}\right)+k\pi =-\frac{5\pi }{12}\pm \frac{1}{2}\arccos \frac{1}{4}+k\pi ,k\in \mathbb{Z}.$$