Bài 3: Một số phương trình lượng giác thường gặp

1.

ĐKXĐ: \(cos3x\ne1\)

\(\dfrac{sin3x}{cos3x-1}=0\)

\(\Rightarrow sin3x=0\)

\(\Leftrightarrow sin^23x=0\)

\(\Leftrightarrow cos^23x=1\)

\(\Rightarrow\left[{}\begin{matrix}cos3x=1\left(loại\right)\\cos3x=-1\end{matrix}\right.\)

\(\Rightarrow3x=\pi+k2\pi\)

\(\Rightarrow x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\)

2.

ĐKXĐ: \(tanx\ne-\sqrt{3}\Rightarrow x\ne-\dfrac{\pi}{3}+k\pi\)

\(\dfrac{sin2x-1}{tanx+\sqrt{3}}=0\)

\(\Rightarrow sin2x-1=0\)

\(\Leftrightarrow sin2x=1\)

\(\Rightarrow2x=\dfrac{\pi}{2}+k2\pi\)

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

giúp tui mấy câu này với

\(\Leftrightarrow3sin^2x-cos^2x-\left(\sqrt{3}sinx-cosx\right)=0\)

\(\Leftrightarrow\left(\sqrt{3}sinx-cosx\right)\left(\sqrt{3}sinx+cosx\right)-\left(\sqrt{3}sinx-cosx\right)=0\)

\(\Leftrightarrow\left(\sqrt{3}sinx-cosx\right)\left(\sqrt{3}sinx+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3}sinx=cosx\\\sqrt{3}sinx+cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}tanx=\dfrac{1}{\sqrt{3}}\\\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\left(\dfrac{\pi}{6}\right)\\sin\left(x+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{6}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

Câu 11: A
Câu 12: B

Đề y/c j bạn nhỉ?

1.

\(sinx+sin5x+sin3x=0\)

\(\Leftrightarrow2sin3x.cos2x+sin3x=0\)

\(\Leftrightarrow sin3x\left(2cos2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

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

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

2.

\(cos7x-cos3x+sin8x+sin2x=0\)

\(\Leftrightarrow-2sin5x.sin2x+2sin5x.cos3x=0\)

\(\Leftrightarrow sin5x\left(sin2x-cos3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin5x=0\\cos3x=sin2x=cos\left(\dfrac{\pi}{2}-2x\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{5}\\x=\dfrac{\pi}{10}+\dfrac{k2\pi}{5}\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

`a)4sin^2 -cos^2 x+sin x cos x=2`

`@TH1:cos x =0<=>x=\pi/2+k\pi` `(k in ZZ)` 

  Ptr có dạng: `4sin^2 x=2<=>sin^2 x=1/2` (Vô lí)

`@TH2:cos x \ne 0<=>x \ne \pi/2+k\pi`

 `=>4tan^2 x-1+tan x=2+2tan^2 x`

`<=>2tan^2 x+tan x-3=0`

`<=>[(tan x=1),(tan x=-3/2):}<=>[(x=\pi/4+k\pi),(x=arc tan(-3/2)^2+k\pi):}`  `(k in ZZ)`  (t/m)

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`b)2sin^2 x-\sqrt{3}sin 2x=2`

`<=>2[1-cos 2x]/2-\sqrt{3}sin 2x=2`

`<=>1-cos 2x-\sqrt{3}sin 2x=2`

`<=>\sqrt{3}sin 2x+cos 2x=-1`

`<=>\sqrt{3}/2sin 2x+1/2cos 2x=-1/2`

`<=>sin (2x+\pi/6)=-1/2`

`<=>[(2x+\pi/6=-\pi/6+k2\pi),(2x+\pi/6=[7\pi]/6+k2\pi):}`

`<=>[(x=-\pi/3+k\pi),(x=\pi/2+k\pi):}`

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`c)2sin 2x+2sin^2 x=3`

`<=>4sin x cos x+2sin^2 x=3`

`@TH1:cos x=0<=>x=\pi/2+k\pi` `(k in ZZ)`

 Ptr có dạng: `2sin^2 x=3<=>sin^2 x=3/2` (Vô lí)

`@TH2:cos x \ne 0<=>x \ne \pi/2+k\pi`  `(k in ZZ)`

  `=>4tan x+2tan^2 x=3+3tan^2 x`

`<=>tan^2 x-4tan x+3=0`

`<=>[(tan x=1),(tan x=3):}<=>[(x=\pi/4+k\pi),(x=arc tan(3)+k\pi):}`  `(k in ZZ)`  (t/m)

`a)3sin^2 x+8sin x.cos x+4cos^2 x=0`

`@TH1:cos x =0<=>x=\pi/2+k\pi` `(k in ZZ)`

  Ptr có dạng: `3sin^2 x=0<=>sin^2 x=0` (Vô lí)

`@TH2:cos x \ne 0<=>x \ne \pi/2+k\pi` `(k in ZZ)`

  `=>3tan^2 x+8tan x+4=0`

`<=>[(tan x=-2/3),(tan x=-2):}<=>[(x=arc tan(-2/3)+k\pi),(x=arc tan(-2)+k\pi):}`  `(k in ZZ)` (t/m)

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`b)sin^2 2x-4sin 2x cos 2x+3cos^2 2x=0`

`@TH1:cos 2x=0<=>x=\pi/4+k\pi/2`  `(k in ZZ)`

 Ptr có dạng: `sin^2 2x=0` (Vô lí)

`@TH2:cos 2x \ne 0<=>x \ne \pi/4+k\pi/2`   `(k ịn ZZ)`

  `=>tan^2 2x-4tan 2x+3=0`

`<=>[(tan 2x=1),(tan 2x=3):}<=>[(2x=\pi/4+k\pi),(2x=arc tan(3)+k\pi):}`

                 `<=>[(x=\pi/8+k\pi/2),(x=[arc tan(3)]/2+k\pi/2):}`   `(k in ZZ)`  (t/m)

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`c)cos^2 x+3sin^2 x+2\sqrt{3}sin x cos x=1`

`@TH1:cos x=0<=>x=\pi/2+k\pi`  `(k in ZZ)`

  Ptr có dạng: `3sin^2 x=1<=>sin^2 x=1/3` (Vô lí)

`@TH2:cos x \ne 0<=>x \ne \pi/2+k\pi`  `(k in ZZ)`

  `=>1+3tan^2 x+2\sqrt{3}tan x=1+tan^2 x`

`<=>2tan^2 x+2\sqrt{3} tan x=0`

`<=>[(tan x =0),(tan x=-\sqrt{3}):}<=>[(x=k\pi),(x=-\pi/3+k\pi):}`   `(k in ZZ)`  (t/m)

a.

\(sin2x-\sqrt{3}cos2x=\sqrt{3}sinx+cosx\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=x+\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\dfrac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

b.

\(sin3x-cos2x=\sqrt{3}\left(sin2x+cos3x\right)\)

\(\Leftrightarrow sin3x-\sqrt{3}cos3x=\sqrt{3}sin2x+cos2x\)

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt[]{3}}{2}cos3x=\dfrac{\sqrt{3}}{2}sin2x+\dfrac{1}{2}cos2x\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=2x+\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\dfrac{5\pi}{6}-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\dfrac{7\pi}{30}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

c.

\(\sqrt{3}cosx-sinx+2sin\left(x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin\left(x-\dfrac{\pi}{3}\right)\)

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

\(\Leftrightarrow x=\dfrac{13\pi}{24}+k\pi\)