Giải các phương trình lượng giác sau:

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

b) $\cos \left(3x+\frac{\pi }{6}\right)=-\frac{\sqrt{2}}{2}$.

c) $\tan (x-15^{\circ })=\sqrt{3}$.

d) $\cot \left(2x+\frac{\pi }{4}\right)=-1$.

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

f) $\cos (2x+30^{\circ })=\sin (x-60^{\circ })$.

(AI giải đề cho tôi)
a) $\sin \left(2x-\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}\iff 2x-\frac{\pi }{3}=\frac{\pi }{3}+k2\pi$ hoặc $2x-\frac{\pi }{3}=\frac{2\pi }{3}+k2\pi$. Suy ra $x=\frac{\pi }{3}+k\pi$ hoặc $x=\frac{\pi }{2}+k\pi$, $k\in \mathbb{Z}$. (Kiểm tra: $x=\frac{\pi }{3}$ thì $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ đúng.)

b) $\cos \left(3x+\frac{\pi }{6}\right)=-\frac{\sqrt{2}}{2}=\cos \frac{3\pi }{4}\iff 3x+\frac{\pi }{6}=\pm \frac{3\pi }{4}+k2\pi$.
• Dấu $+$: $3x=\frac{3\pi }{4}-\frac{\pi }{6}+k2\pi =\frac{7\pi }{12}+k2\pi \iff x=\frac{7\pi }{36}+\frac{2k\pi }{3}$.
• Dấu $-$: $3x=-\frac{3\pi }{4}-\frac{\pi }{6}+k2\pi =-\frac{11\pi }{12}+k2\pi \iff x=-\frac{11\pi }{36}+\frac{2k\pi }{3}$.
Vậy $x=\frac{7\pi }{36}+\frac{2k\pi }{3}$ hoặc $x=-\frac{11\pi }{36}+\frac{2k\pi }{3}$, $k\in \mathbb{Z}$. (Kiểm tra: $x=-\frac{11\pi }{36}$ thì $3x+\frac{\pi }{6}=-\frac{3\pi }{4}$, $\cos \left(-\frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}$ đúng.)

c) Điều kiện $x-15^{\circ }\ne 90^{\circ }+k180^{\circ }$. $\tan (x-15^{\circ })=\sqrt{3}=\tan 60^{\circ }\iff x-15^{\circ }=60^{\circ }+k180^{\circ }\iff x=75^{\circ }+k180^{\circ }$, $k\in \mathbb{Z}$ (thỏa điều kiện).

d) Điều kiện $2x+\frac{\pi }{4}\ne k\pi$. $\cot \left(2x+\frac{\pi }{4}\right)=-1=\cot \left(-\frac{\pi }{4}\right)\iff 2x+\frac{\pi }{4}=-\frac{\pi }{4}+k\pi \iff 2x=-\frac{\pi }{2}+k\pi \iff x=-\frac{\pi }{4}+\frac{k\pi }{2}$, $k\in \mathbb{Z}$ (thỏa điều kiện vì $2x+\frac{\pi }{4}=k\pi$).

e) $\sin 3x+\sin \left(x-\frac{\pi }{4}\right)=0\iff \sin 3x=\sin \left(\frac{\pi }{4}-x\right)$
$\iff 3x=\frac{\pi }{4}-x+k2\pi$ hoặc $3x=\pi -\left(\frac{\pi }{4}-x\right)+k2\pi =\frac{3\pi }{4}+x+k2\pi$
$\iff x=\frac{\pi }{16}+\frac{k\pi }{2}$ hoặc $x=\frac{3\pi }{8}+k\pi$, $k\in \mathbb{Z}$.

f) $\sin (x-60^{\circ })=\cos (90^{\circ }-(x-60^{\circ }))=\cos (150^{\circ }-x)$. Phương trình trở thành $\cos (2x+30^{\circ })=\cos (150^{\circ }-x)$
$\iff 2x+30^{\circ }=\pm (150^{\circ }-x)+k360^{\circ }$.
• Dấu $+$: $3x=120^{\circ }+k360^{\circ }\iff x=40^{\circ }+k120^{\circ }$.
• Dấu $-$: $2x+30^{\circ }=-150^{\circ }+x+k360^{\circ }\iff x=-180^{\circ }+k360^{\circ }$.
Vậy $x=40^{\circ }+k120^{\circ }$ hoặc $x=-180^{\circ }+k360^{\circ }$, $k\in \mathbb{Z}$. (Kiểm tra: $x=40^{\circ }$: $\cos 110^{\circ }=-\sin 20^{\circ }=\sin (-20^{\circ })=\sin (-20^{\circ })$ đúng.)