Giải các phương trình sau

a) $\tan \left(\frac{x}{2}+\frac{\pi }{6}\right)=\sqrt{3}$

b) $\sin \left(3x+\frac{\pi }{4}\right)=\frac{\sqrt{3}}{2}$.

c) $\cos \left(2x-70^{\circ }\right)=-1$.

d) $3\sin \left(-2x+38^{\circ }\right)=4$.

e) $\sin \left(3x-\frac{7\pi }{12}\right)=\sin \left(-x+\frac{\pi }{4}\right)$.

f) $\cos \left(x+\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}-x\right)=0$.

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

(AI giải đề cho tôi)
Bài 5 (Phần 4): Giải từng phương trình lượng giác.

a) $\tan \left(\frac{x}{2}+\frac{\pi }{6}\right)=\sqrt{3}$
Điều kiện: $\frac{x}{2}+\frac{\pi }{6}\ne \frac{\pi }{2}+k\pi$.
$\frac{x}{2}+\frac{\pi }{6}=\frac{\pi }{3}+k\pi \Rightarrow \frac{x}{2}=\frac{\pi }{6}+k\pi \Rightarrow x=\frac{\pi }{3}+2k\pi ,k\in \mathbb{Z}$ (thỏa điều kiện).

b) $\sin \left(3x+\frac{\pi }{4}\right)=\frac{\sqrt{3}}{2}$
Trường hợp 1: $3x+\frac{\pi }{4}=\frac{\pi }{3}+2k\pi \Rightarrow 3x=\frac{\pi }{12}+2k\pi \Rightarrow x=\frac{\pi }{36}+\frac{2k\pi }{3}$.
Trường hợp 2: $3x+\frac{\pi }{4}=\frac{2\pi }{3}+2k\pi \Rightarrow 3x=\frac{5\pi }{12}+2k\pi \Rightarrow x=\frac{5\pi }{36}+\frac{2k\pi }{3}$, với $k\in \mathbb{Z}$.

c) $\cos (2x-70^{\circ })=-1$
$2x-70^{\circ }=180^{\circ }+k360^{\circ }\Rightarrow x=125^{\circ }+k180^{\circ },k\in \mathbb{Z}$.

d) $3\sin (-2x+38^{\circ })=4\Rightarrow \sin (-2x+38^{\circ })=\frac{4}{3}>1$: vô nghiệm vì $\sin \le 1$.

e) $\sin \left(3x-\frac{7\pi }{12}\right)=\sin \left(-x+\frac{\pi }{4}\right)$
Trường hợp 1: $3x-\frac{7\pi }{12}=-x+\frac{\pi }{4}+2k\pi \Rightarrow 4x=\frac{5\pi }{6}+2k\pi \Rightarrow x=\frac{5\pi }{24}+\frac{k\pi }{2}$.
Trường hợp 2: $3x-\frac{7\pi }{12}=\pi -\left(-x+\frac{\pi }{4}\right)+2k\pi =x+\frac{3\pi }{4}+2k\pi \Rightarrow 2x=\frac{4\pi }{3}+2k\pi \Rightarrow x=\frac{2\pi }{3}+k\pi$, với $k\in \mathbb{Z}$.

f) $\cos \left(x+\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}-x\right)=0$
Dùng công thức tổng hai cos: $2\cos \frac{(x+\pi /4)+(\pi /4-x)}{2}\cos \frac{(x+\pi /4)-(\pi /4-x)}{2}=2\cos \frac{\pi }{4}\cos x=\sqrt{2}\cos x=0$
$\Rightarrow x=\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}$.

g) $\sin 3x-\cos \left(\frac{3\pi }{4}-x\right)=0\iff \sin 3x=\cos \left(\frac{3\pi }{4}-x\right)=\sin \left(\frac{\pi }{2}+\frac{3\pi }{4}-x\right)=\sin \left(\frac{5\pi }{4}-x\right)$
Trường hợp 1: $3x=\frac{5\pi }{4}-x+2k\pi \Rightarrow 4x=\frac{5\pi }{4}+2k\pi \Rightarrow x=\frac{5\pi }{16}+\frac{k\pi }{2}$.
Trường hợp 2: $3x=\pi -\left(\frac{5\pi }{4}-x\right)+2k\pi =-\frac{\pi }{4}+x+2k\pi \Rightarrow 2x=-\frac{\pi }{4}+2k\pi \Rightarrow x=-\frac{\pi }{8}+k\pi$, với $k\in \mathbb{Z}$.