Bài 3: Một số phương trình lượng giác thường gặp

2: Đặt sinx+cosx=t

\(\Leftrightarrow1+sin2x=t^2\)

=>sin 2x=t²-1

Theo đề, ta có: \(2t+3t^2-3-2=0\)

\(\Leftrightarrow3t^2+2t-5=0\)

=>t=1 hoặc t=-5/3

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=1\\sinx+cosx=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\dfrac{\Pi}{4}\right)=1\\\sqrt{2}sin\left(x+\dfrac{\Pi}{4}\right)=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{\sqrt{2}}\\sin\left(x+\dfrac{\Pi}{4}\right)=-\dfrac{5}{3\sqrt{2}}\left(loại\right)\end{matrix}\right.\)

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

`a)6cos^2 x+5sin x-7=0`

`<=>6(1-sin^2 x)+5sin x-7=0`

`<=>-6sin^2 x+5sin x-1=0`

Đặt `sin x=t` `(-1 <= t <= 1)` 

  `=>-6t^2+5t-1=0`

`<=>t=1/2` hoặc `t=1/3`

       (t/m)                 (t/m)

`@t=1/2=>sin x=1/2<=>` $\left[\begin{matrix} x=\dfrac{\pi}{6}+k2\pi\\ x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.$   `(k in ZZ)`

`@t=1/3=>sin x=1/3<=>` $\left[\begin{matrix} x=arc sin(\dfrac{1}{3})+k2\pi\\ x=\pi-arc sin(\dfrac{1}{3})+k2\pi\end{matrix}\right.$  `(k in ZZ)`

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`b)3sin^2 2x+7cos 2x-3=0`

`<=>3(1-cos^2 2x)+7cos 2x-3=0`

`<=>-3cos^2 2x+7cos 2x=0`

Đặt `cos 2x=t`  `(-1 <= t <= 1)`

   `=>-3t^2+7t=0`

`<=>t=7/3` hoặc `t=0`

     (ko t/m)              (t/m)

  `@t=0=>cos 2x=0<=>2x=\pi/2+k\pi<=>x=\pi/4+[k\pi]/2`   `(k in ZZ)`

P/t \(\Leftrightarrow2cos2x.cosx-2.sin2x.cos2x=0\)

\(\Leftrightarrow cos2x\left(cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=1\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{2}+k\pi\\x=2k\pi\end{matrix}\right.\)  ( k \(\in Z\) )

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

\(\Leftrightarrow2\sqrt{3}\cdot sin2x-2sin2x+4cos2x=\dfrac{5}{2}\)

\(\Leftrightarrow sin2x\cdot\left(2\sqrt{3}-2\right)+4cos2x=\dfrac{5}{2}\)

\(\Leftrightarrow sin2x\left(4\sqrt{3}-4\right)+8cos2x=5\)

Đến đây là phương trình cơ bản theo dạng \(a\cdot sinx+b\cdot cosx=c\) rồi, bạn chỉ cần làm theo dạng đó là xong

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

`@TH1:cos x=0=>x=\pi/2+k\pi`     `(cos x=0=>sin^2 x=1)`

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

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

   `=>3([sin x]/[cos x])^2+8 [sin x]/[cos x]+4=0`

`<=>3tan^2 x+8tan x+4=0`

`<=>` $\left[\begin{matrix} tan x=\dfrac{-2}{3}\\ tan x=-2\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=arc tan(\dfrac{-2}{3})+k\pi\\ x=arc tan(-2)+k\pi\end{matrix}\right.$   `(k in ZZ)`   (t/m)

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`b)sin^2 x-8sin x cos x+4cos^2 x=0`

`@TH1:cos x=0<=>x=\pi/2+k\pi`    `(cos x=0<=>sin^2 x=1)`

  Ptr có dạng: `sin^2 x=0` (Vô lí)

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

   `=>([sin x]/[cos x])^2-8[sin x]/[cos x]+4=0`

`<=>tan^2 x-8 tan x+4=0`

`<=>` $\left[\begin{matrix} tan x=4+2\sqrt{3}\\ tan x=4-2\sqrt{3}\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=arc tan(4+2\sqrt{3})+k\pi\\ x=arc tan(4-2\sqrt{3})+k\pi\end{matrix}\right.$   `(k in ZZ)`   (t/m)

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

`@TH1:cos x=0<=>x=\pi/2+k\pi`    `(cos x=0<=>sin^2 x=1)`

  Ptr có dạng: `-sin^2 x=3` (Vô lí)

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

    `=>4+3[sin x]/[cos x]-([sin x]/[cos x])^2=3/[cos^2 x]`

`<=>4+3tan x-tan^2 x=3(1+tan^2 x)`

`<=>4tan^2 x-3tan x-1=0`

`<=>` $\left[\begin{matrix} tan x=1\\ tan x=\dfrac{-1}{4}\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{\pi}{4}+k\pi\\ x=arc tan(\dfrac{-1}{4})+k\pi\end{matrix}\right.$    `(k in ZZ)`   (t/m)

a: \(\Leftrightarrow3\cdot\dfrac{1-cos2x}{2}+4\cdot sin2x+4\cdot\dfrac{1+cos2x}{2}=0\)

\(\Leftrightarrow-\dfrac{3}{2}cos2x+\dfrac{3}{2}+4sin2x+2+2cos2x=0\)

\(\Leftrightarrow4sin2x+\dfrac{1}{2}cos2x=-\dfrac{7}{2}\)

\(\Leftrightarrow8sin2x+cos2x=-7\)

\(\Leftrightarrow sin2x\cdot\dfrac{8}{\sqrt{65}}+cos2x\cdot\dfrac{1}{\sqrt{65}}=\dfrac{-7}{\sqrt{65}}\)

Đặt \(cosa=\dfrac{8}{\sqrt{65}}\)

Pt sẽ là \(sin\left(2x+a\right)=\dfrac{-7}{65}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+a=arcsin\left(-\dfrac{7}{65}\right)+k2\Pi\\2x+a=\Pi-arcsin\left(-\dfrac{7}{65}\right)+k2\Pi\end{matrix}\right.\)

=>...

b: \(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)-4sin2x+4\cdot\dfrac{1+cos2x}{2}=0\)

\(\Leftrightarrow-\dfrac{1}{2}cos2x-4sin2x+\dfrac{1}{2}+2+2cos2x=0\)

\(\Leftrightarrow-4sin2x+\dfrac{3}{2}cos2x=-\dfrac{5}{2}\)

\(\Leftrightarrow8sin2x-3cos2x=5\)

\(\Leftrightarrow sin2x\cdot\dfrac{8}{\sqrt{73}}-cos2x\cdot\dfrac{3}{\sqrt{73}}=\dfrac{5}{\sqrt{73}}\)

\(\Leftrightarrow sin\left(2x-a\right)=\dfrac{5}{\sqrt{73}}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-a=arcsin\left(\dfrac{5}{\sqrt{73}}\right)+k2\Pi\\2x-a=\Pi-arcsin\left(\dfrac{5}{\sqrt{73}}\right)+k2\Pi\end{matrix}\right.\)

=>...

f: \(\Leftrightarrow\tan x-\dfrac{1}{\tan x}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{tan^2x-1}{tanx}=\dfrac{3}{2}\)

\(\Leftrightarrow2tan^2x-2-3tanx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=2\\tanx=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(2\right)+k\Pi\\x=arctan\left(-\dfrac{1}{2}\right)+k\Pi\end{matrix}\right.\)

i: \(\Delta=\left(\sqrt{3}-1\right)^2-4\cdot\left(\sqrt{3}\right)=4-2\sqrt{3}+4\sqrt{3}=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)

Pt có hai nghiệm phân biệt là:

\(\left[{}\begin{matrix}tanx=\dfrac{-\sqrt{3}+1-\sqrt{3}-1}{2}=-\sqrt{3}\\tanx=\dfrac{-\sqrt{3}+1+\sqrt{3}+1}{2}=1\end{matrix}\right.\)

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

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

\(\Leftrightarrow cos^2x-cosx-2=0\)

\(\Leftrightarrow cosx=-1\)

hay \(x=\Pi+k2\Pi\)

d: \(\Leftrightarrow2cos^2x-1+9cosx+5=0\)

\(\Leftrightarrow2cos^2x+9cosx+4=0\)

\(\Leftrightarrow2cos^2x+8cosx+cosx+4=0\)

\(\Leftrightarrow cosx=-\dfrac{1}{2}\)

hay \(x=\pm\dfrac{2\Pi}{3}+k2\Pi\)

a: \(\Leftrightarrow1-2sin^2x-3sinx+2=0\)

\(\Leftrightarrow2sin^2x-3sinx-3=0\)

\(\Leftrightarrow sinx=\dfrac{3-\sqrt{33}}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\dfrac{3-\sqrt{33}}{4}\right)+k2\Pi\\x=\Pi-arcsin\left(\dfrac{3-\sqrt{33}}{4}\right)+k2\Pi\end{matrix}\right.\)

g: \(\Leftrightarrow1-sin^2x+sinx+1=0\)

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

\(\Leftrightarrow sinx=-1\)

hay \(x=-\dfrac{\Pi}{2}+k2\Pi\)

`a)cos 2x-3sin x+2=0`

`<=>1-2sin^2 x-3sin x+2=0`

`<=>` $\left[\begin{matrix} sin x=\dfrac{-3+\sqrt{33}}{4}\\ sin x=\dfrac{-3-\sqrt{33}}{4} (VN)\end{matrix}\right.$

`<=>sin x=[-3+\sqrt{33}]/4`

`<=>` $\left[\begin{matrix} x=arc sin\dfrac{-3+\sqrt{33}}{4}+k2\pi\\ x=\pi-arc sin\dfrac{-3+\sqrt{33}}{4}+k2\pi\end{matrix}\right.$   `(k in ZZ)`

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`c)4 sin^2 2x-8cos^2 x+3=0`

`<=>4 sin^2 2x-(8cos^2 x -4)-1=0`

`<=>4(1-cos^2 2x)-4 cos 2x-1=0`

`<=>4-4cos^2 2x-4 cos 2x-1=0`

`<=>` $\left[\begin{matrix} cos x=\dfrac{1}{2}\\ cos x=\dfrac{-3}{2}(VN)\end{matrix}\right.$

`<=>cos x=1/2`

`<=>x=[+-\pi]/3+k2\pi`   `(k in ZZ)`

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`e)cot^2` `x/2+4cot` `x/2+3=0`

`<=>` $\left[\begin{matrix} cot \dfrac{x}{2}=-1\\ cot \dfrac{x}{2}=-3\end{matrix}\right.$

`<=>` $\left[\begin{matrix} \dfrac{x}{2}=\dfrac{-\pi}{4}+k\pi\\ \dfrac{x}{2}=arc cot (-3)+k\pi\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{-\pi}{2}+k2\pi\\ x=2arc cot(-3)+k2\pi\end{matrix}\right.$    `(k in ZZ)`

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`g)cos^2 x+sin x+1=0`

`<=>1-sin^2 x+sin x+1=0`

`<=>` $\left[\begin{matrix} sin x=-1\\ sin x=-2 (VN)\end{matrix}\right.$

`<=>sin x=-1`

`<=>x=[-\pi]/2+k2\pi`    `(k in ZZ)`

1) ĐK : \(x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

 \(...\Leftrightarrow1+cot^2x=cotx+3\)  \(\Leftrightarrow cot^2x-cotx-2=0\)  

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

Đến đây ; bn làm tiếp

2) \(...\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[sin\left(4x-\dfrac{\pi}{2}\right)+sin2x\right]=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[-cos4x+sin2x\right]=0\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}\left[2sin^22x-1+sin2x\right]=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(sin^22x+sin2x+1\right)=0\) \(\Leftrightarrow PTVN\)

mn giúp em vs ạ