Đáp án D

\(\Leftrightarrow sin3x.cosx+cos3x.sinx-2\left(sin^23x+cos^23x\right)+cos3x=0\)

\(\Leftrightarrow sin4x+cos3x-2=0\)

Do \(\left\{{}\begin{matrix}sin4x\le1\\cos3x\le1\end{matrix}\right.\) \(\Rightarrow sin4x+cos3x-2\le0\)

Dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sin4x=1\\cos3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=\frac{\pi}{2}+k2\pi\\3x=n2\pi\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{2}\\x=\frac{n2\pi}{3}\end{matrix}\right.\)

Biểu diễn trên đường tròn lượng giác thì 2 tập nghiệm này ko có điểm chung

Vậy pt vô nghiệm

1.

\(\Leftrightarrow cos3x=-\frac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=40^0+k120^0\\x=-40^0+k120^0\end{matrix}\right.\)

\(\Rightarrow x=\left\{40^0;160^0;80^0\right\}\)

2.

Bạn coi lại đề, số \(-\sqrt{3}\) bên vế trái ko hề hợp lý, toán cho cấp 1 như vầy còn được chứ cấp 3 chắc ko ai cho đề kiểu vậy đâu

3.

\(\Leftrightarrow\sqrt{3}sin3x-cos3x=-sin5x-\sqrt{3}cos5x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=-\left(\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\right)\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(-5x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=-5x-\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\frac{4\pi}{3}+5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{48}+\frac{k\pi}{4}\\x=-\frac{7\pi}{12}+k\pi\end{matrix}\right.\)

a/

\(\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=sin\left(x+\frac{\pi}{6}\right)\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)

\(\Rightarrow x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi\)

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

b/

\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin\frac{\pi}{12}\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=sin\frac{\pi}{12}\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{12}+k2\pi\\x+\frac{\pi}{6}=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos3x\)

\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos3x\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)

d/

\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=sin2x\)

\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{15}+\frac{k2\pi}{5}\end{matrix}\right.\)

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

\(\Leftrightarrow3\sqrt{3}\sin x-4\sqrt{3}\sin^3x-2\sin x.\cos x=0\)

Xét \(\sin x=0\) là nghiệm của pt=> \(x=\pi\)

Xét \(\sin x\ne0\) chia cả 2 vế cho \(\sin x\)

\(\Leftrightarrow3\sqrt{3}-4\sqrt{3}\sin^2x-2\cos x=0\)

\(\Leftrightarrow3\sqrt{3}-4\sqrt{3}+4\sqrt{3}\cos^2x-2\cos x=0\)

\(\Leftrightarrow4\sqrt{3}\cos^2x-2\cos x-\sqrt{3}=0\)

đây là pt b2, bn tự giải và kết luận :3

Đề là \(\sqrt{5}cos4x\) hay \(\sqrt{3}cos4x\) bạn?

\(\Leftrightarrow\sqrt{3}\sin3x=\sin2x \)

\(\Rightarrow3\sin^23x=\sin^22x\)

\(\Leftrightarrow\frac{3}{2}\left(1-\cos6x\right)=1-\cos^22x\)

\(\Leftrightarrow\frac{3}{2}\left[1-\left(4\cos^32x-3\cos2x\right)\right]=1-\cos^22x\)

đến đây đặt ẩn cos2x rồi giải tiếp nhé cậu ^^

Lời giải:

PT \(\Leftrightarrow \frac{\sqrt{3}}{2}(3\sin x-4\sin ^3x)-\sin x\cos x=0\)

\(\Leftrightarrow \sin x(3\sqrt{3}-4\sin ^2x-2\cos x)=0\)

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

\(\Rightarrow \left[\begin{matrix} \sin x=0\\ 4\cos ^2-2\cos x+3\sqrt{3}-4=0\end{matrix}\right.\)

Nếu \(\sin x=0\Rightarrow x=k\pi \) với $k$ nguyên bất kỳ

Nếu \(4\cos ^2x-2\cos x+3\sqrt{3}-4=0\)

\(\Leftrightarrow (2\cos x-\frac{1}{2})^2=\frac{17-12\sqrt{3}}{4}< 0\) (vô lý- loại)

Vậy............

VP `=2sin (3x)/2 . cos (3x)/2`

`=sin (2. (3x)/2)`

`=sin3x=` VT.

`(sin2a=2sinacosa)`

\(\Leftrightarrow cos5x=-sin3x\)

\(\Leftrightarrow cos5x=cos\left(\dfrac{\pi}{2}+3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}+3x+k2\pi\\5x=-\dfrac{\pi}{2}-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{16}+\dfrac{k\pi}{4}\end{matrix}\right.\)