a/

\(\Leftrightarrow sinx=2cosx\)

Nhận thấy \(cosx=0\) không phải nghiệm, pt tương đương:

\(\frac{sinx}{cosx}=2\Leftrightarrow tanx=2\)

\(\Leftrightarrow tanx=tana\) (với \(a\in\left(0;\frac{\pi}{2}\right)\) sao cho \(tana=2\))

\(\Rightarrow x=a+k\pi\)

b/

\(tan2x=cotx=tan\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow2x=\frac{\pi}{2}-x+k\pi\)

\(\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{3}\)

c/

\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x=0\)

\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx-\sqrt{3}cosx=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\sinx=\sqrt{3}cosx\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\Leftrightarrow tanx=\sqrt{3}\)

\(\Rightarrow x=\frac{\pi}{3}+k\pi\)

d/

\(\Leftrightarrow tan\left(3x-50^0\right)=-cot\left(x-30^0\right)\)

\(\Leftrightarrow tan\left(3x-50^0\right)=tan\left(x+60^0\right)\)

\(\Rightarrow3x-50^0=x+60^0+k180^0\)

\(\Rightarrow x=55^0+k90^0\)

1. \(sin\left(\dfrac{\pi}{3}-x\right)\ne0\Leftrightarrow\dfrac{\pi}{3}-x\ne k\pi\Leftrightarrow x\ne\dfrac{\pi}{3}-k\pi\)

2. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

3. \(\sqrt{1+sinx}-\sqrt{2}\ge0\Leftrightarrow1+sinx\ge2\Leftrightarrow sinx\ge1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

4. \(\sqrt{2-2cosx}-2\ne0\Leftrightarrow2-2cosx\ne4\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\)

5. \(1-\sqrt{1+sin3x}\ne0\Leftrightarrow sin3x\ne0\Leftrightarrow3x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{3}\)

1/ ĐKXĐ: \(cos2x\ne0\Rightarrow2x\ne k\frac{\pi}{2}\Rightarrow x\ne\frac{k\pi}{4}\)

2/ ĐKXĐ:

\(\sqrt{2-2cosx}\ne2\Rightarrow2-2cosx\ne4\)

\(\Rightarrow cosx\ne-1\Rightarrow x\ne\pi+k2\pi\)

3/ ĐKXĐ: \(sin3x\ne0\Rightarrow3x\ne k\pi\Rightarrow x\ne\frac{k\pi}{3}\)

ĐKXĐ: ...

\(\Leftrightarrow\frac{6tanx}{1-tan^2x}-3tanx-\frac{5}{2}=0\)

\(\Leftrightarrow12tanx-6tanx\left(1-tan^2x\right)-5=0\)

\(\Leftrightarrow6tan^3x+6tanx-5=0\)

Bạn coi lại đề :)

Chọn  A

a/ \(sinx=-\frac{\sqrt{3}}{2}=sin\left(-\frac{\pi}{3}\right)\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

b/ \(cosx=\frac{\sqrt{3}}{2}=cos\left(\frac{\pi}{6}\right)\Rightarrow x=\pm\frac{\pi}{6}+k2\pi\)

c/ \(cosx=\frac{\sqrt{2}}{2}=cos\left(\frac{\pi}{4}\right)\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)

d/ \(tanx=-\sqrt{3}=tan\left(-\frac{\pi}{3}\right)\Rightarrow x=-\frac{\pi}{3}+k\pi\)

a/

ĐKXĐ: ...

\(\Leftrightarrow tanx-8\sqrt{3}=3tanx-6\sqrt{3}\)

\(\Leftrightarrow2tanx=-2\sqrt{3}\)

\(\Rightarrow tanx=-\sqrt{3}\Rightarrow x=-\frac{\pi}{3}+k\pi\)

b/

\(\Leftrightarrow tan2x=-cot\left(\frac{5\pi}{8}\right)\)

\(\Leftrightarrow tan2x=tan\left(\frac{\pi}{2}+\frac{5\pi}{8}\right)\)

\(\Leftrightarrow tan2x=tan\left(\frac{9\pi}{8}\right)\)

\(\Rightarrow2x=\frac{9\pi}{8}+k\pi\Rightarrow x=\frac{9\pi}{16}+\frac{k\pi}{2}\)

c/

\(\Leftrightarrow\sqrt{3}tan\left(\frac{\pi}{9}-2x\right)=-3\)

\(\Leftrightarrow tan\left(\frac{\pi}{9}-2x\right)=-\sqrt{3}\)

\(\Rightarrow\frac{\pi}{9}-2x=-\frac{\pi}{3}+k\pi\)

\(\Rightarrow x=\frac{2\pi}{9}+\frac{k\pi}{2}\)

d/

\(\Leftrightarrow\left[{}\begin{matrix}tanx=5\\tan2x=tan4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\2x=4+k\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(5\right)+k\pi\\x=2+\frac{k\pi}{2}\end{matrix}\right.\)