1.

ĐK: \(x\ne\dfrac{\pi}{4}+k\pi\)

\(\dfrac{cos2x}{1-sin2x}=0\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Đối chiếu điều kiên ta được \(x=-\dfrac{\pi}{4}+k\pi\)

2.

ĐK: \(x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3};x\ne\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

\(tan3x=tan4x\)

\(\Leftrightarrow3x=4x+k\pi\)

\(\Leftrightarrow x=k\pi\)

3.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(cot2x.sin3x=0\)

\(\Leftrightarrow\dfrac{cos2x}{sin2x}.sin3x=0\)

\(\Leftrightarrow cos2x.sin3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\3x=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{3}\end{matrix}\right.\)

Đáp án B

Đáp án B

a.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)

\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)

b.

\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

c.

\(\Leftrightarrow\dfrac{1}{2}cos6x+\dfrac{1}{2}cos4x=\dfrac{1}{2}cos6x+\dfrac{1}{2}cos2x+4-3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(2cos^22x-1\right)=\dfrac{1}{2}cos2x+\dfrac{5}{2}+\dfrac{3}{2}cos2x\)

\(\Leftrightarrow cos^22x-2cos2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)

a/

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x+\frac{1}{2}cos3x=\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\)

\(\Leftrightarrow sin\left(3x+\frac{\pi}{6}\right)=sin\left(x+\frac{\pi}{3}\right)\)

\(\Rightarrow\left[{}\begin{matrix}3x+\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\3x+\frac{\pi}{6}=\pi-x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

\(\Leftrightarrow2\left(\frac{1+cos2x}{2}\right)-3\sqrt{3}sin2x-4\left(\frac{1-cos2x}{2}\right)=-4\)

\(\Leftrightarrow3cos2x-3\sqrt{3}sin2x=-3\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x=1\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

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

c/

Ủa đề câu này bạn ghi đúng ko? Nhìn kì kì, cos8x hay cos3x bên vế phải vậy?

d/

\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{3}=x-\frac{\pi}{3}+k2\pi\\2x+\frac{\pi}{3}=\frac{\pi}{3}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2\pi}{3}+k2\pi\\x=\frac{k2\pi}{3}\end{matrix}\right.\)

e/

\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)

\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)

\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)

Ta có:

Vậy giá trị của biểu thức  không phụ thuộc vào x.