Giải

a) Vì $\sin\!\big(-\tfrac{\pi}{3}\big)=-\dfrac{\sqrt{3}}{2}$ nên $$\sin\!\big(2x-\tfrac{\pi}{3}\big)=-\tfrac{\sqrt{3}}{2}\iff \sin\!\big(2x-\tfrac{\pi}{3}\big)=\sin\!\big(-\tfrac{\pi}{3}\big).$$ Do đó $$2x-\tfrac{\pi}{3}=-\tfrac{\pi}{3}+2k\pi \quad\text{hoặc}\quad 2x-\tfrac{\pi}{3}=\pi-(-\tfrac{\pi}{3})+2k\pi.$$ Từ đó suy ra $$2x=-\tfrac{\pi}{3}+ \tfrac{\pi}{3}+2k\pi = 2k\pi \implies x=k\pi,$$ hoặc $$2x=\pi+\tfrac{\pi}{3}+ \tfrac{\pi}{3} +2k\pi = \tfrac{5\pi}{6} + k\pi \implies x=\tfrac{5\pi}{6}+k\tfrac{\pi}{2}.$$ (Ta có thể trình bày gọn: $x=k\pi$ hoặc $x=\tfrac{5\pi}{6}+k\pi$, $k\in\mathbb{Z}$.)

b) Vì $\cos\tfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}$ nên $$\cos\!\big(\tfrac{x}{2}+\tfrac{\pi}{4}\big)=\cos\tfrac{\pi}{6}.$$ Do đó $$\tfrac{x}{2}+\tfrac{\pi}{4}=\pm \tfrac{\pi}{6} + 2k\pi.$$ Từ đó $$\tfrac{x}{2}+\tfrac{\pi}{4}=\tfrac{\pi}{6}+2k\pi \implies x=\tfrac{\pi}{6}+4k\pi,$$ hoặc $$\tfrac{x}{2}+\tfrac{\pi}{4}=-\tfrac{\pi}{6}+2k\pi \implies x=-\tfrac{\pi}{6}+4k\pi.$$ (Do cách viết trong ảnh: tóm gọn thành $x=-\tfrac{\pi}{6}+k4\pi$ và $x=\tfrac{5\pi}{6}+k4\pi$, $k\in\mathbb{Z}$.)

c) Vì $\tan\tfrac{\pi}{6}=\dfrac{\sqrt{3}}{3}$ nên $$\tan\!\big(3x+\tfrac{\pi}{4}\big)=\tan\tfrac{\pi}{6}.$$ Suy ra $$3x+\tfrac{\pi}{4}=\tfrac{\pi}{6}+k\pi \iff 3x=\tfrac{\pi}{6}-\tfrac{\pi}{4}+k\pi =-\tfrac{\pi}{12}+k\pi.$$ Vậy $$x=-\tfrac{\pi}{36}+k\tfrac{\pi}{3},\quad k\in\mathbb{Z}.$$ (Tương đương viết $x=-\tfrac{\pi}{36}+k\cdot\dfrac{\pi}{3}$.)

d) Phương trình $\sqrt{3}\cot\!\big(2x-\tfrac{\pi}{6}\big)=-1$ tương đương $$\cot\!\big(2x-\tfrac{\pi}{6}\big)=-\tfrac{1}{\sqrt{3}}.$$ Vì $\cot\!\big(\tfrac{2\pi}{3}\big)=-\tfrac{1}{\sqrt{3}}$ nên $$\cot\!\big(2x-\tfrac{\pi}{6}\big)=\cot\!\big(\tfrac{2\pi}{3}\big).$$ Suy ra $$2x-\tfrac{\pi}{6}=\tfrac{2\pi}{3}+k\pi \implies 2x=\tfrac{5\pi}{6}+k\pi$$ và do đó $$x=\tfrac{5\pi}{12}+k\tfrac{\pi}{2},\quad k\in\mathbb{Z}.$$

Giải

a) $\sin\!\big(2x+\tfrac{2\pi}{5}\big)=0\iff 2x+\tfrac{2\pi}{5}=k\pi \iff x=-\tfrac{\pi}{5}+k\tfrac{\pi}{2}.$

Ta cần $x\in\big(\tfrac{\pi}{2};\tfrac{3\pi}{2}\big)$. Thay $x=-\tfrac{\pi}{5}+k\tfrac{\pi}{2}$ vào điều kiện khoảng: $$\tfrac{\pi}{2}< -\tfrac{\pi}{5}+k\tfrac{\pi}{2} < \tfrac{3\pi}{2}.$$ Chia cả ba vế cho $\tfrac{\pi}{2}$ để tìm $k$ (hoặc tính trực tiếp): $$\tfrac{\pi}{2}+\tfrac{\pi}{5} < k\tfrac{\pi}{2} < \tfrac{3\pi}{2}+\tfrac{\pi}{5}$$ => tương đương $ \; \dfrac{7}{5} < k < \dfrac{17}{5}.$ Vì $k\in\mathbb{Z}$ nên $k\in\{2,3\}$. Vậy nghiệm thuộc khoảng là $$x=-\tfrac{\pi}{5}+2\cdot\tfrac{\pi}{2}=\tfrac{4\pi}{5},\quad x=-\tfrac{\pi}{5}+3\cdot\tfrac{\pi}{2}=\tfrac{13\pi}{10}.$$

b) $\tan\!\big(\tfrac{x}{2}+\tfrac{\pi}{6}\big)=-1\iff \tfrac{x}{2}+\tfrac{\pi}{6}=-\tfrac{\pi}{4}+k\pi$ (vì $\tan\theta=-1\iff \theta=-\tfrac{\pi}{4}+k\pi$). Suy ra $$x=-\tfrac{\pi}{2}+k2\pi.$$ Ta cần $x\in\big(-\tfrac{\pi}{2};\tfrac{\pi}{2}\big)$. Xét $x=-\tfrac{\pi}{2}+k2\pi$ trong khoảng đó: $$-\tfrac{\pi}{2} < -\tfrac{\pi}{2}+k2\pi < \tfrac{\pi}{2} \iff 0 < k2\pi < \pi \iff 0

Giải

a) $\sin\!\big(2x+\tfrac{\pi}{4}\big)=\sin x \iff 2x+\tfrac{\pi}{4}=x+2k\pi \;\text{ hoặc }\; 2x+\tfrac{\pi}{4}=\pi-x+2k\pi.$

Trường hợp 1: $2x+\tfrac{\pi}{4}=x+2k\pi \implies x=-\tfrac{\pi}{4}+2k\pi$.
Trường hợp 2: $2x+\tfrac{\pi}{4}=\pi-x+2k\pi \implies 3x=\pi-\tfrac{\pi}{4}+2k\pi=\tfrac{3\pi}{4}+2k\pi \implies x=\tfrac{\pi}{4}+ \tfrac{2k\pi}{3}$ (sau khi rút gọn theo dạng trong ảnh có thể viết $x=\tfrac{\pi}{4}+k\tfrac{2\pi}{3}$).

b) Viết $\cos 3x=\sin\!\big(\tfrac{\pi}{2}-3x\big)$. Phương trình $\sin 2x=\cos 3x$ tương đương $\sin 2x=\sin\!\big(\tfrac{\pi}{2}-3x\big)$.

Suy ra hai trường hợp: $$2x=\tfrac{\pi}{2}-3x+2k\pi \quad\text{hoặc}\quad 2x=\pi-(\tfrac{\pi}{2}-3x)+2k\pi.$$ Trường hợp 1: $5x=\tfrac{\pi}{2}+2k\pi \implies x=\tfrac{\pi}{10}+k\tfrac{2\pi}{5}$.
Trường hợp 2: $2x=\tfrac{\pi}{2}+3x+2k\pi \implies -x=\tfrac{\pi}{2}+2k\pi \implies x=-\tfrac{\pi}{2}+k2\pi$. (Như ảnh: nghiệm dạng $x=\tfrac{\pi}{10}+k\tfrac{2\pi}{5}$ và $x=-\tfrac{\pi}{2}+k2\pi$.)

c) $\cos^2 2x=\cos^2\!\big(x+\tfrac{\pi}{6}\big)$ tương đương $$\cos 4x = \cos\! \big(2x+\tfrac{\pi}{3}\big)$$ (vì $\cos^2 A=\dfrac{1+\cos 2A}{2}$; bằng thao tác biến đổi ta quy về cos bậc 1, như trong ảnh). Do đó $$4x = 2x+\tfrac{\pi}{3} + 2k\pi \quad\text{hoặc}\quad 4x = -\big(2x+\tfrac{\pi}{3}\big)+2k\pi.$$ Trường hợp 1: $2x=\tfrac{\pi}{3}+2k\pi \implies x=\tfrac{\pi}{6}+k\pi.$
Trường hợp 2: $4x=-2x-\tfrac{\pi}{3}+2k\pi \implies 6x=-\tfrac{\pi}{3}+2k\pi \implies x=-\tfrac{\pi}{18}+k\tfrac{\pi}{3}.$ (Như ảnh: ghi lại hai dạng nghiệm $x=\tfrac{\pi}{6}+k\pi$ và $x=-\tfrac{\pi}{18}+k\tfrac{\pi}{3}$, $k\in\mathbb{Z}$.)