a/

\(\Leftrightarrow4cos^3x-3cosx+2cos^2x-1-cosx-1=0\)

\(\Leftrightarrow2cos^3x+cos^2x-2cosx-1=0\)

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

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

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

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

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

b/

\(cos\left(8sinx\right)=1\)

\(\Leftrightarrow8sinx=k2\pi\)

\(\Leftrightarrow sinx=\frac{k\pi}{4}\)

Do \(-1\le sinx\le1\Rightarrow-1\le\frac{k\pi}{4}\le1\)

\(\Rightarrow k=\left\{-1;0;1\right\}\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{\pi}{4}\\sinx=0\\sinx=\frac{\pi}{4}\end{matrix}\right.\)

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

c/

\(\Leftrightarrow1+2cos^2x-1+cosx=0\)

\(\Leftrightarrow2cos^2x-cosx=0\)

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

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

d/

Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\ge0\\cosx=b\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}a+3b=2\\a^2+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2-3b\\a^2+b^2=1\end{matrix}\right.\)

\(\Rightarrow\left(2-3b\right)^2+b^2-1=0\)

\(\Rightarrow10b^2-12b+3=0\Rightarrow\left[{}\begin{matrix}b=\frac{6+\sqrt{6}}{10}\Rightarrow a=\frac{2-3\sqrt{6}}{10}\left(l\right)\\b=\frac{6-\sqrt{6}}{10}\Rightarrow a=\frac{2+3\sqrt{6}}{10}\end{matrix}\right.\)

\(\Rightarrow cosx=\frac{6-\sqrt{6}}{10}\)

\(\Rightarrow x=\pm arccos\left(\frac{6-\sqrt{6}}{10}\right)+k2\pi\)

b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)

\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)

\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)

c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)

\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)

a. cos2x + cos4x + cos6x = 0

\(\Leftrightarrow\left(cos2x+cos6x\right)+cos4x=0\\ \Leftrightarrow2cos4x.cos2x+cos4x=0\\ \Leftrightarrow cos4x\left(2cos2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\left(k\in Z\right)}\)

1.

\(cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2cos4x.cos2x+cos4x=0\)

\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

2.

\(\Leftrightarrow1+cos2x+cosx+cos3x=0\)

\(\Leftrightarrow1+2cos^2x-1+2cos2x.cosx=0\)

\(\Leftrightarrow cos^2x+cos2x.cosx=0\)

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

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

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

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

\(...\Leftrightarrow cosx+2cos^2x-1+4cos^3x-3cosx=0\left(1\right)\)

Đặt \(cosx=a\left(-1\le a\le1\right)\).

\(\left(1\right)\Leftrightarrow4a^3+2a^2-2a-1=0\)

\(\Leftrightarrow\left(2a+1\right)\left(2a^2-1\right)\Leftrightarrow\left[{}\begin{matrix}a=-\frac{1}{2}\\a=\pm\sqrt{\frac{1}{2}}\end{matrix}\right.\Leftrightarrow...\)

Chọn D

Phương trình tương với:

Trên đường tròn đơn vị, các điểm nghiệm của phương trình là 4 điểm A, B, C, D như hình vẽ. Do đó trên nửa khoảng [ - π ; 0 ) , phương trình có đúng 2 nghiệm (là - π  và - 2 π 3 ).

Đáp án D.

Phương trình tương với:

cos x − 2 cos 2 x − 1 − 4 cos 3 x − 3 cos x + 1 = 0 ⇔ − 4 cos 3 x − 2 cos 2 x + 4 cos x + 2 = 0 ⇔ 2 t 3 + t 2 − − 2 t − 1 = 0 t = cos x ⇔ t 2 − 1 2 t + 1 = 0 ⇔ t = 1 t = − 1 t = − 1 2

Trên đường tròn đơn vị, các điểm nghiệm của phương trình là 4 điểm A, B, C, D như hình vẽ. Do đó trên nửa khoảng − π ; 0 , phương trình có đúng 2 nghiệm (là − π  và − 2 π 3 ).

@Nguyễn Việt Lâm giúp em với ạ

a/ \(4cos^3x-3cosx-4\left(2cos^2x-1\right)+3cosx-4=0\)

\(\Leftrightarrow4cos^3x-8cos^2x=0\)

\(\Leftrightarrow4cos^2x\left(cosx-2\right)=0\)

\(\Leftrightarrow cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(0< \frac{\pi}{2}+k\pi< 14\Rightarrow-\frac{1}{2}< k< \frac{14-\frac{\pi}{2}}{\pi}\Rightarrow k=\left\{0;1;2;3\right\}\)

\(\Rightarrow x=\left\{\frac{\pi}{2};\frac{3\pi}{2};\frac{5\pi}{2};\frac{7\pi}{2}\right\}\)

b/ Bạn coi lại đề, cái ngoặc thứ 2 thiếu \(\left(2cos\left(???\right)+cosx\right)\)

c/ Bạn coi lại đề, có 2 số hạng \(cos2x\) xuất hiện ở vế trái, cấp 3 chắc ko ai cho kiểu vậy đâu, nếu đúng thế thì người ta cộng luôn thành \(2cos2x\) cho rồi

2.

\(sin3x+cos2x=1+2sinx.cos2x\)

\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)

\(\Leftrightarrow cos2x+sinx-1=0\)

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

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

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

1.

\(cos3x-cos4x+cos5x=0\)

\(\Leftrightarrow cos3x+cos5x-cos4x=0\)

\(\Leftrightarrow2cos4x.cosx-cos4x=0\)

\(\Leftrightarrow\left(2cosx-1\right)cos4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)

3.

\(cos2x-cosx=2sin^2\dfrac{3x}{2}\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}+2sin^2\dfrac{3x}{2}=0\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}+sin\dfrac{3x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{3x}{2}.sinx.cos\dfrac{x}{2}=0\)

Đến đây dễ rồi tự làm tiếp nha.