Question

Giải phương trình:

1.\(cos^3x.cos3x+sin^3x.sin3x=\frac{\sqrt{2}}{4}\)

2.\(cos^34x=cos^3x.cos3x+sin^3x.sin3x\)

3.\(cos^2x-4sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)+2=0\)

4.\(sin^4x+sin^4\left(x+\frac{\pi}{4}\right)=\frac{1}{4}\)

5.\(sin^6x+cos^6x=\frac{5}{6}\left(sin^4x+cos^4x\right)\)

6.\(sin^6x+cos^6x+\frac{1}{2}sinx.cosx=0\)

7.\(\frac{1}{2}\left(sin^4x+cos^4x\right)=sin^2x.cos^2x+sinx.cosx\)

8.\(sin^6x+cos^6x-3cos8x+2=0\)

9.\(sin^4x+cos^4x-2\left(sin^6\frac{x}{2}+cos^6\frac{x}{2}\right)+1=0\)

Answer 1

5.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)

\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)

\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)

\(\Leftrightarrow sin^22x=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)

Answer 2

6.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)

\(\Leftrightarrow-3sin^22x+sin2x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)

Answer 3

1.

\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)

\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)

\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)

\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)

\(\Leftrightarrow4cos^32x=\sqrt{2}\)

\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)

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

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

Answer 4

2.

\(\Leftrightarrow4cos^34x=4cos^3x.cos3x+4sin^3x.sin3x\)

\(\Leftrightarrow3cos4x+cos12x=\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x\)

\(\Leftrightarrow3cos4x+cos12x=3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x\)

\(\Leftrightarrow3cos4x+cos12x=3cos2x+cos6x\)

\(\Leftrightarrow3\left(cos4x-cos2x\right)+cos12x-cos6x=0\)

\(\Leftrightarrow3sin3x.sinx+sin9x.sin3x=0\)

\(\Leftrightarrow sin3x\left(3sinx+sin9x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\left(1\right)\\3sinx+sin9x=0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\frac{k\pi}{3}\)

\(\left(2\right)\Leftrightarrow3sinx+3sin3x-4sin^33x=0\)

\(\Leftrightarrow3sinx+9sinx-12sin^3x-4\left(3sinx-4sin^3x\right)^3=0\)

\(\Leftrightarrow12sinx\left(1-sin^2x\right)-4sin^3x\left(3-4sin^2x\right)^3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\12cos^2x-4sin^2x\left(2cos2x+1\right)^3=0\left(3\right)\end{matrix}\right.\)

\(\left(3\right)\Leftrightarrow6cos2x+6-2\left(1-cos2x\right)\left(2cos2x+1\right)^3=0\)

Đặt \(cos2x=t\Rightarrow3t+3-\left(1-t\right)\left(2t+1\right)^3=0\)

\(\Leftrightarrow4t^4+2t^3-3t^2-t+1=0\)

Pt này vô nghiệm

Answer 5

3.

\(cos^2x+2\left[1-2sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)\right]=0\)

\(\Leftrightarrow cos^2x+2cos\left(x-\frac{\pi}{2}\right)=0\)

\(\Leftrightarrow cos^2x+2sinx=0\)

\(\Leftrightarrow-sin^2x+2sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\sqrt{2}+1\left(l\right)\\sinx=\sqrt{2}-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\sqrt{2}-1\right)+k2\pi\\x=\pi-arcsin\left(\sqrt{2}-1\right)+k2\pi\end{matrix}\right.\)

Answer 6

4.

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

\(\Leftrightarrow\left(1-cos2x\right)^2+\left[1-cos\left(2x+\frac{\pi}{2}\right)\right]^2=1\)

\(\Leftrightarrow\left(1-cos2x\right)^2+\left(1+sin2x\right)^2=1\)

\(\Leftrightarrow cos^22x-2cos2x+1+sin^22x+2sin2x+1=1\)

\(\Leftrightarrow sin2x-cos2x=-1\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\2x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

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

Answer 7

7.

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

\(\Leftrightarrow1-2sin^2x.cos^2x=2sin^2x.cos^2x+2sinx.cosx\)

\(\Leftrightarrow4sin^2x.cos^2x+2sinx.cosx-1=0\)

\(\Leftrightarrow sin^22x+sin2x-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{5}-1}{2}\\sin2x=\frac{-\sqrt{5}-1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}arcsin\left(\frac{\sqrt{5}-1}{2}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(\frac{\sqrt{5}-1}{2}\right)+k\pi\end{matrix}\right.\)

Answer 8

8.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-3cos8x+2=0\)

\(\Leftrightarrow1-3sin^2x.cos^2x-3cos8x+2=0\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x-3cos8x+2=0\)

\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)-3\left(2cos^24x-1\right)+2=0\)

\(\Leftrightarrow-48cos^24x+3cos4x+45=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=1\\cos4x=-\frac{15}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=k2\pi\\4x=arccos\left(-\frac{15}{16}\right)+k2\pi\\4x=-arccos\left(-\frac{15}{16}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{1}{4}arccos\left(-\frac{15}{16}\right)+\frac{k\pi}{2}\\x=-\frac{1}{4}arccos\left(-\frac{15}{16}\right)+\frac{k\pi}{2}\end{matrix}\right.\)

Answer 9

9.

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-2\left[\left(sin^2\frac{x}{2}+cos^2\frac{x}{2}\right)^3-3sin^2\frac{x}{2}.cos^2\frac{x}{2}\right]+1=0\)

\(\Leftrightarrow1-2sin^2x.cos^2x-2+6sin^2\frac{x}{2}.cos^2\frac{x}{2}+1=0\)

\(\Leftrightarrow-2sin^2x.cos^2x+\frac{3}{2}sin^2x=0\)

\(\Leftrightarrow sin^2x\left(3-4cos^2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\frac{\sqrt{3}}{2}\\cosx=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{\pi}{6}+k2\pi\\x=\pm\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)