Trắc nghiệm thì chuyển hết sang vế trái, sau đó cho đại x 1 giá trị nào đó ko đẹp (ví dụ \(\frac{\pi}{5}\)) rồi dùng tính năng CALC để bấm, cái nào ra bằng 0 thì chọn (chọn x ko đẹp để loại trừ khả năng tình cờ đúng ở các giá trị đặc biệt)

Còn ko thì biến đổi từng con một:

a/ \(sin^4x-cos^4x=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)=sin^2x-cos^2x\)

\(=1-cos^2x-cos^2x=1-2cos^2x\) (đúng luôn)

Khỏi cần quan tâm các câu còn lại

\(\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=sin4x\)

\(\Leftrightarrow cos^2x-sin^2x=sin4x\)

\(\Leftrightarrow cos2x=sin4x=cos\left(\frac{\pi}{2}-4x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-4x+k2\pi\\2x=4x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{3}\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(2cos^2x-1=sin6x\)

\(\Leftrightarrow cos2x=sin6x=cos\left(\frac{\pi}{2}-6x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-6x+k2\pi\\2x=6x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{16}+\frac{k\pi}{4}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

\(2\left(cos^2x-1\right)=sinx.cos3x\)

\(\Leftrightarrow-2sin^2x=sinx.cos3x\)

\(\Leftrightarrow sinx.cos3x+2sin^2x=0\)

\(\Leftrightarrow sinx\left(cos3x+2sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos3x+2sinx=0\left(1\right)\end{matrix}\right.\)

Bạn có ghi nhầm đề ko nhỉ, pt (1) dù giải được nhưng khá khó đấy, phải vận dụng công thức nhân 3 và nghiệm ko hề đẹp

\(cos^4x+sin^4x=1+\frac{1}{2}sin4x\)

\(\Leftrightarrow\left(cos^2x+sin^2x\right)^2-2\left(sinx.cosx\right)^2=1+\frac{1}{2}sin4x\)

\(\Leftrightarrow1-\frac{1}{2}sin^22x=1+\frac{1}{2}sin4x\)

\(\Leftrightarrow sin4x+sin^22x=0\)

\(\Leftrightarrow2sin2x.cos2x+sin^22x=0\)

\(\Leftrightarrow sin2x\left(2cos2x+sin2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=0\Rightarrow x=\frac{k\pi}{2}\\2cos2x+sin2x=0\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x+\frac{2}{\sqrt{5}}cos2x=0\)

Đặt \(cosa=\frac{1}{\sqrt{5}}\) với \(a\in\left[0;\pi\right]\)

\(\Rightarrow sin2x.cosa+cos2x.sina=0\)

\(\Leftrightarrow sin\left(2x+a\right)=0\)

\(\Rightarrow2x+a=k\pi\Rightarrow x=-\frac{a}{2}+\frac{k\pi}{2}\)

a/

\(\Leftrightarrow\left(sin^2\frac{x}{3}+cos^2\frac{x}{3}\right)^2-2sin^2\frac{x}{3}.cos^2\frac{x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}sin^2\frac{2x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{4}\left(1-cos\frac{4x}{3}\right)=\frac{5}{8}\)

\(\Leftrightarrow cos\frac{4x}{3}=-\frac{1}{2}\)

\(\Leftrightarrow\frac{4x}{3}=\pm\frac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\frac{\pi}{2}+\frac{k3\pi}{2}\)

b/

\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow4-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\)

\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-1\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x+\frac{1}{2}cos4x=-\frac{1}{2}\)

\(\Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)

c/

\(\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)

\(\Leftrightarrow-cos^32x+5cos^22x-7cos2x+3=0\)

\(\Leftrightarrow\left(3-cos2x\right)\left(cos2x-1\right)^2=0\)

\(\Leftrightarrow cos2x=1\)

\(\Leftrightarrow x=k\pi\)

d/

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos4x\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=cos4x\)

\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)=cos4x\)

\(\Leftrightarrow cos4x=1\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

a/

\(cos^6x+sin^2x=1\)

\(\Leftrightarrow cos^6x-\left(1-sin^2x\right)=0\)

\(\Leftrightarrow cos^6x-cos^2x=0\)

\(\Leftrightarrow cos^2x\left(cos^4x-1\right)=0\)

\(\Leftrightarrow cos^2x\left(cos^2x-1\right)\left(cos^2x+1\right)=0\)

\(\Leftrightarrow-cos^2x.sin^2x=0\)

\(\Leftrightarrow sin^22x=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

b/

\(cos^6x-sin^6x=\frac{13}{18}cos^22x\)

\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(cos^4x+sin^4x+sin^2x.cos^2x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left[\left(sin^2x+cos^2x\right)^2-sin^2x.cos^2x\right]=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(1-\frac{1}{4}sin^22x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(1-\frac{1}{4}\left(1-cos^22x\right)\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(\frac{3}{4}+\frac{1}{4}cos^22x\right)=\frac{13}{18}cos^22x\)

\(\Leftrightarrow cos2x\left(\frac{1}{4}cos^22x-\frac{13}{18}cos2x+\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\\frac{1}{4}cos^22x-\frac{13}{18}cos2x+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

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

c/

\(cos^4x+sin^6x=cos2x\)

\(\Leftrightarrow\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)

\(\Leftrightarrow cos^32x-5cos^2x+7cos2x-3=0\)

\(\Leftrightarrow\left(cos2x-1\right)^2\left(cos2x-3\right)=0\)

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

\(\Leftrightarrow2x=k2\pi\)

\(\Rightarrow x=k\pi\)

$M=si{n}^{4}x\left(1+2co{s}^{2}x\right)+co{s}^{4}x\left(1+2si{n}^{2}x\right)$

$=\; Sin4x+2Sin4x.cos2x+cos4x+2sin2x.cos4x$

$=\; sin4x+cos4x\; +2sin2x.cos2x(sin2x+cos2x)$

$=\; sin4x+cos4x+2sin2x.cos2x\; =(sin2x+cos2x)2=12=1$

\(B=cos^2x.cot^2x+cos^2x-cot^2x+2\left(sin^2x+cos^2x\right)\)

\(=cos^2x\left(cot^2x+1\right)-cot^2x+2\)

\(=\frac{cos^2x}{sin^2x}-cot^2x+1=cot^2x-cot^2x+1=1\)

\(M=cos^4x-sin^4x+cos^4x+sin^2x.cos^2x+3sin^2x\)

\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)

\(=cos^2x-sin^2x+cos^2x+3sin^2x\)

\(=2\left(sin^2x+cos^2x\right)=2\)

Giả sử các biểu thức đều có nghĩa

\(A=2\left(\left(sin^2x\right)^3+\left(cos^2x\right)^3\right)-3\left(sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x\right)\)

\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)

\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)

\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)

b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)

\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)

c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)

\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)

\(C=cos^2x-sin^2x+cos^2x+3sin^2x\)

\(C=2cos^2x+2sin^2x=2\left(cos^2x+sin^2x\right)=2\)