Question

1. Cosx.cos7x = cos3x.cos5x

2. Sin3x - 4sinx.cos2x = 0

3. Sinx + 3sin2x + sin3x = 0

4. Sinbình2x - cosbình2x + 3/4 = 0

5. Sinbìnhx + sinbình2x = 1

Answer 1

1.

\(\Leftrightarrow\frac{1}{2}\left(cos8x+cos6x\right)=\frac{1}{2}\left(cos8x+cos2x\right)\)

\(\Leftrightarrow cos6x=cos2x\)

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

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

2.

\(sin3x-4sinx.cos2x=0\)

\(\Leftrightarrow sin3x-2\left(sin3x-sinx\right)=0\)

\(\Leftrightarrow-sin3x+2sinx=0\)

\(\Leftrightarrow4sin^3x-3sinx+2sinx=0\)

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

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

Answer 2

5.

\(sin^2x+sin^22x=1\)

\(\Leftrightarrow4sin^2x.cos^2x-\left(1-sin^2x\right)=0\)

\(\Leftrightarrow4sin^2x.cos^2x-cos^2x=0\)

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

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

\(\Rightarrow x=...\)

Answer 3

3.

\(sinx+sin3x+3sin2x=0\)

\(\Leftrightarrow2sin2x.cosx+3sin2x=0\)

\(\Leftrightarrow sin2x\left(2cosx+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cosx=-\frac{3}{2}< -1\left(l\right)\end{matrix}\right.\)

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

Answer 4

4.

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

\(\Leftrightarrow-cos4x+\frac{3}{4}=0\)

\(\Leftrightarrow cos4x=\frac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}4x=arccos\left(\frac{3}{4}\right)+k2\pi\\4x=-arccos\left(\frac{3}{4}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}arccos\left(\frac{3}{4}\right)+k2\pi\\x=-\frac{1}{4}arccos\left(\frac{3}{4}\right)+k2\pi\end{matrix}\right.\)

Answer 5

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

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

\(\Leftrightarrow-cos^22x-\left(2cos^2x-1\right)+\frac{3}{4}=0\)

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

Đặt \(cos2x=t\) \(\Rightarrow\left|t\right|\le1\)

\(\Rightarrow-t^2-t+\frac{3}{4}=0\Rightarrow\left[{}\begin{matrix}t=\frac{1}{2}\\t=-\frac{3}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cos2x=\frac{1}{2}\)

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

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