Bài 2: Phương trình lượng giác cơ bản

`cos 4x=cos 12x`

`<=>` $\left[\begin{matrix} 4x=12x+k2\pi\\ 4x=-12x+k2\pi\end{matrix}\right.$  `(k in ZZ)`

`<=>` $\left[\begin{matrix} 8x=-k2\pi\\ 16x=k2\pi\end{matrix}\right.$  `(k in ZZ)`

`<=>` $\left[\begin{matrix} x=\dfrac{-k\pi}{4}\\ x=\dfrac{k\pi}{8}\end{matrix}\right.$  `(k in ZZ)`

\(\Leftrightarrow\left\{{}\begin{matrix}4x=12x+k2\Pi\\4x=-12x+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-k\Pi}{8}\\x=\dfrac{k\Pi}{16}\end{matrix}\right.\)

\(\Leftrightarrow\sin\left(\dfrac{\Pi}{4}-x\right)=\sin\left(\dfrac{\Pi}{2}-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{-\Pi}{4}+k2\Pi\\-3x=\dfrac{1}{4}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{4}-k2\Pi\\x=\dfrac{-1}{12}\Pi-\dfrac{2}{3}k\Pi\end{matrix}\right.\)

e: =>sin 2x=-sin(x+pi/3)

=>sin2x=sin(-x-pi/3)

=>2x=-x-pi/3+k2pi hoặc 2x=x+4/3pi+k2pi

=>x=4/3pi+k2pi hoặc x=-pi/9+k2pi/3

g: =>sin(pi/4-x)=sin(pi/2-2x)

=>pi/4-x=pi/2-2x+k2pi hoặc pi/4-x=pi/2+2x+k2pi

=>x=pi/4+k2pi hoặc -3x=pi/4+k2pi

=>x=pi/4+k2pi hoặc x=-pi/12-k2pi/3

Đề bài là: \(sin^2\left(\dfrac{\pi}{8}+a\right)-sin^2\left(\dfrac{\pi}{8}-a\right)-\dfrac{\sqrt{2}}{2}sin2a\) đúng không nhỉ?

\(=\dfrac{1}{2}-\dfrac{1}{2}cos\left(\dfrac{\pi}{4}+2a\right)-\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{\pi}{4}-2a\right)-\dfrac{\sqrt{2}}{2}sin2a\)

\(=\dfrac{1}{2}\left[cos\left(\dfrac{\pi}{4}-2a\right)-cos\left(\dfrac{\pi}{4}+2a\right)\right]-\dfrac{\sqrt{2}}{2}sin2a\)

\(=sin\left(\dfrac{\pi}{4}\right).sin2a-\dfrac{\sqrt{2}}{2}sin2a=\dfrac{\sqrt{2}}{2}sin2a-\dfrac{\sqrt{2}}{2}sin2a=0\)

P/t \(\Leftrightarrow2cos2x.sin2x-sin2x+2cos^22x-cos2x-1=0\)

\(\Leftrightarrow sin4x-sin2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sinx.cos3x-2sin3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(1\right)\\cos3x=sin3x\left(2\right)\end{matrix}\right.\) 

(1) \(\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) \(\Leftrightarrow sin3x-cos3x=0\)  \(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\left(k\in Z\right)\)

Vậy ... 

Với \(sinx=0\) không phải nghiệm (vế trái bằng 9, vế phải hiển nhiên nhỏ hơn 9)

Với \(sinx\ne0\):

\(\Rightarrow\left(3sinx-4sin^3x\right)\left(3-4sin^23x\right)=sinx-2sinx.cos10x\)

\(\Leftrightarrow sin3x\left(3-4sin^23x\right)=sinx-2sinx.cos10x\)

\(\Leftrightarrow3sin3x-4sin^33x=sinx-sin11x+sin9x\)

\(\Leftrightarrow sin9x=sinx-sin11x+sin9x\)

\(\Leftrightarrow sin11x=sinx\)

\(\Leftrightarrow...\)

22.

\(sinx\le1\Rightarrow3sinx+2019\le3.1+2019=2022\)

Đáp án B

23.

\(C_n^3=\dfrac{A_n^3}{3!}=\dfrac{2022}{6}=337\)

Đáp án A