Question

1)\(sin^23x.cos2x+sin^2x=0\)

2)

\(cos^23x+cos^22x=sin^2x\)

3)

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

4)

\(sin^23x-sin^22x-sin^2x=0\)

5)

\(2cos^2x=3sin^25x+2\)

6) 3cosx+2cos2x-cos3x=2sinxsin2x-1

7) \(sinx+cosx=\sqrt{2}\left(2-sin^32x\right)\)

Answer 1

1.

\(\Leftrightarrow\left(1-cos6x\right)cos2x+1-cos2x=0\)

\(\Leftrightarrow cos2x-cos2x.cos6x+1-cos2x=0\)

\(\Leftrightarrow\frac{1}{2}\left(cos8x-cos4x\right)-1=0\)

\(\Leftrightarrow2cos^24x-cos4x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\cos4x=\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow4x=\pi+k2\pi\)

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

Answer 2

2.

\(\Leftrightarrow1+cos6x+2cos^22x=1-cos2x\)

\(\Leftrightarrow cos6x+cos2x+2cos^22x=0\)

\(\Leftrightarrow cos4x.cos2x+cos^22x=0\)

\(\Leftrightarrow cos2x\left(cos4x+cos2x\right)=0\)

\(\Leftrightarrow cos2x\left(2cos^22x+cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

Answer 3

3.

Đặt \(\frac{x}{6}=t\Rightarrow\frac{1}{4}+cos^22t=\frac{1}{2}sin^23t\)

\(\Leftrightarrow1+4cos^22t=1-cos6t\)

\(\Leftrightarrow cos6t+4cos^22t=0\)

\(\Leftrightarrow4cos^32t+4cos^22t-3cos2t=0\)

\(\Leftrightarrow cos2t\left(4cos^22t+4cos2t-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2t=0\\cos2t=\frac{1}{2}\\cos2t=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{4}+\frac{k\pi}{2}\\t=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{x}{3}=\frac{\pi}{6}+k\pi\\\frac{x}{3}=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow x=...\)

Answer 4

4.

\(1-cos6x-2sin^22x-\left(1-cos2x\right)=0\)

\(\Leftrightarrow cos2x-cos6x-2sin^22x=0\)

\(\Leftrightarrow sin4x.sin2x-sin^22x=0\)

\(\Leftrightarrow2sin^22x.cos2x-sin^22x=0\)

\(\Leftrightarrow sin^22x\left(2cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)

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

Answer 5

5.

\(\Leftrightarrow2\left(cos^2x-1\right)=3sin^25x\)

\(\Leftrightarrow-2sin^2x=3sin^25x\)

Do \(\left\{{}\begin{matrix}VT=-2sin^2x\le0\\VP=3sin^25x\ge0\end{matrix}\right.\) ; \(\forall x\)

Nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sinx=0\\sin5x=0\end{matrix}\right.\) \(\Leftrightarrow sinx=0\)

\(\Leftrightarrow x=k\pi\)

Answer 6

6.

\(3cosx+2cos2x-cos3x=cosx-cos3x-1\)

\(\Leftrightarrow2cosx+2\left(2cos^2x-1\right)+1=0\)

\(\Leftrightarrow4cos^2x+2cosx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{-1+\sqrt{5}}{4}\\cosx=\frac{-1-\sqrt{5}}{4}\end{matrix}\right.\)

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

7.

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\left(2-sin^32x\right)\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)+sin^32x=2\)

Do \(\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)\le1\\sin^32x\le1\end{matrix}\right.\) ; \(\forall x\) nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=1\\sin2x=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

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