a.

\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)

\(tan3x-tanx=0\)

\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)

\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)

\(\Leftrightarrow sin2x=0\)

\(\Leftrightarrow2sinx.cosx=0\)

\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)

c.

\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)

\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)

\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

d.

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

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

\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)

a) cosx - √3sinx = √2 ⇔ cosx - tansinx = √2

⇔ coscosx - sinsinx = √2cos ⇔ cos(x + ) =

⇔

b) 3sin3x - 4cos3x = 5 ⇔ sin3x - cos3x = 1.

Đặt α = arccos thì phương trình trở thành

cosαsin3x - sinαcos3x = 1 ⇔ sin(3x - α) = 1 ⇔ 3x - α = + k2π

⇔ x = , k ∈ Z (trong đó α = arccos).

c/

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

\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\)

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

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

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

a/

\(\Leftrightarrow\sqrt{3}cos2x-\left(sin^2x+cos^2x-2sinx.cosx\right)=2\)

\(\Leftrightarrow\sqrt{3}cos2x-1+sin2x=2\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x=\frac{3}{2}\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{3}\right)=\frac{3}{2}\)

Vế phải lớn hơn 1 nên pt vô nghiệm

b/

\(\Leftrightarrow\frac{5}{2}\left(1+cos2x\right)+2sin2x=4\)

\(\Leftrightarrow4sin2x+5cos2x=3\)

\(\Leftrightarrow\frac{4}{\sqrt{41}}sin2x+\frac{5}{\sqrt{41}}cos2x=\frac{3}{\sqrt{41}}\)

Đặt \(\frac{4}{\sqrt{41}}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sin2x.cosa+cos2x.sina=\frac{3}{\sqrt{41}}\)

\(\Leftrightarrow sin\left(2x+a\right)=\frac{3}{\sqrt{41}}=sinb\)

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

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

.

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.

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

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

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

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

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