Bài 2: Phương trình lượng giác cơ bản

H24

Giải phương trình

NL
16 tháng 9 2022 lúc 16:17

18.

\(1-sin^22x+2\left(sinx+cosx\right)^3-3sin2x-3=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^3-\left(sin^22x+3sin2x+2\right)=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^3-\left(sin2x+1\right)\left(sin2x+2\right)=0\)

\(\Leftrightarrow2\left(sinx+cosx\right)^3+2\left(sinx+cosx\right)^2\left(sinx.cosx+1\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)^2\left(sinx+cosx+sinx.cosx+1\right)=0\)

\(\Leftrightarrow\left(sinx+cosx\right)^2\left(sinx+1\right)\left(cosx+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\sinx=-1\\cosx=-1\end{matrix}\right.\)

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

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NL
16 tháng 9 2022 lúc 16:21

19.

\(\left(cosx-sinx\right)\left(cos^2x+sin^2x+sinx.cosx\right)=1\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(1+sinx.cosx\right)=1\)

Đặt \(cosx-sinx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow t^2=1-2sinx.cosx\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

Pt trở thành:

\(t\left(1+\dfrac{1-t^2}{2}\right)=1\)

\(\Leftrightarrow t^3-3t+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow cosx-sinx=1\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

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NL
16 tháng 9 2022 lúc 16:27

20.

\(sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x+1=2cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\)

\(\Leftrightarrow sin\dfrac{x}{2}.sinx-cos\dfrac{x}{2}.sin^2x-\left[2cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)-1\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}.sinx-cos\dfrac{x}{2}.sin^2x-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}sinx-cos\dfrac{x}{2}sin^2x-sinx=0\)

\(\Leftrightarrow sinx\left(sin\dfrac{x}{2}-cos\dfrac{x}{2}.sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sin\dfrac{x}{2}-cos\dfrac{x}{2}sinx-1=0\left(1\right)\end{matrix}\right.\)

Xét (1) 

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}cos^2\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\left(1-sin^2\dfrac{x}{2}\right)-1=0\)

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\)

\(\Leftrightarrow...\)

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NL
16 tháng 9 2022 lúc 16:29

21.

\(4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\)

\(\Leftrightarrow4\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]+\sqrt{3}sin4x=2\)

\(\Leftrightarrow4-2sin^22x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-1\)

\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{\sqrt{3}}{2}sin4x=-\dfrac{1}{2}\)

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

\(\Leftrightarrow...\)

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NL
16 tháng 9 2022 lúc 16:38

22.

\(sin4x-cos4x=1+4\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow1-2sin2x.cos2x+cos^22x-sin^22x+4\left(sinx-cosx\right)=0\)

\(\Leftrightarrow\left(cos2x-sin2x\right)^2+\left(cos2x-sin2x\right)\left(cos2x+sin2x\right)+4\left(sinx-cosx\right)=0\)

\(\Leftrightarrow2cos2x\left(cos2x-sin2x\right)+4\left(sinx-cosx\right)=0\)

\(\Leftrightarrow\left(cos^2x-sin^2x\right)\left(cos2x-sin2x\right)+2\left(sinx-cosx\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)\left(cos2x-sin2x\right)-2\left(cosx-sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=sinx\Leftrightarrow x=...\\\left(cosx+sinx\right)\left(cos2x-sin2x\right)-2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cosx.cos2x-cosx.sin2x+sinx.cos2x-sinx.sin2x-2=0\)

\(\Leftrightarrow\left(cosx.cos2x-sinx.sin2x\right)+\left(sinx.cos2x-cosx.sin2x\right)=2\)

\(\Leftrightarrow cos3x-sinx=2\)

\(\Leftrightarrow\left(cos3x-1\right)+\left(1-sinx\right)=0\)

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

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NL
16 tháng 9 2022 lúc 16:40

23.

\(cos7x-sin5x=\sqrt{3}\left(cos5x-sin7x\right)\)

\(\Leftrightarrow\sqrt{3}sin7x+cos7x=sin5x+\sqrt{3}cos5x\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin7x+\dfrac{1}{2}cos7x=\dfrac{1}{2}sin5x+\dfrac{\sqrt{3}}{2}cos5x\)

\(\Leftrightarrow sin\left(7x+\dfrac{\pi}{6}\right)=sin\left(5x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}7x+\dfrac{\pi}{6}=5x+\dfrac{\pi}{3}+k2\pi\\7x+\dfrac{\pi}{6}=\dfrac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

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NL
16 tháng 9 2022 lúc 16:49

25.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(1+cot2x=\dfrac{1-cos2x}{sin^22x}\)

\(\Leftrightarrow\dfrac{sin2x+cos2x}{sin2x}=\dfrac{1-\left(1-2sin^2x\right)}{4sin^2x.cos^2x}\)

\(\Leftrightarrow\dfrac{sin2x+cos2x}{sinx.cosx}=\dfrac{1}{cos^2x}\)

\(\Rightarrow sin2x.cosx+cos2x.cosx=sinx\)

\(\Leftrightarrow sin3x+sinx+cos3x+cosx=2sinx\)

\(\Leftrightarrow sin3x+cos3x=sinx-cosx\)

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

\(\Leftrightarrow...\)

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NL
16 tháng 9 2022 lúc 16:54

26.

\(sin3x+cos2x=1-2sinx.cos2x\)

\(\Leftrightarrow3sinx-4sin^3x+1-2sin^2x=1-2sinx\left(1-2sin^2x\right)\)

\(\Leftrightarrow3sinx-4sin^3x+1-2sin^2x=1-2sinx+4sin^3x\)

\(\Leftrightarrow8sin^3x+2sin^2x-5sinx=0\)

\(\Leftrightarrow sinx\left(8sin^2x+2sinx-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{-1+\sqrt{41}}{8}\\sinx=\dfrac{-1-\sqrt{41}}{8}\end{matrix}\right.\)

\(\Leftrightarrow...\)

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NL
16 tháng 9 2022 lúc 16:58

27.

\(sin^2x+sin^23x-3cos^22x=0\)

\(\Leftrightarrow1-cos2x+1-cos6x-3\left(1+cos4x\right)=0\)

\(\Leftrightarrow3cos4x+cos6x+cos2x+1=0\)

\(\Leftrightarrow3cos4x+2cos4x.cos2x+1=0\)

\(\Leftrightarrow3\left(2cos^22x-1\right)+2\left(2cos^22x-1\right).cos2x+1=0\)

\(\Leftrightarrow2cos^32x+3cos^22x-cos2x-1=0\)

\(\Leftrightarrow\left(2cos2x+1\right)\left(cos^22x+cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-\dfrac{1}{2}\\cos2x=\dfrac{-1-\sqrt{5}}{2}< -1\left(loại\right)\\cos2x=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

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NL
16 tháng 9 2022 lúc 17:04

28.

ĐKXĐ: \(x\ne\left\{\dfrac{\pi}{2}+k2\pi;-\dfrac{\pi}{6}+k2\pi;\dfrac{7\pi}{6}+k2\pi\right\}\)

\(\dfrac{\left(1-2sinx\right)cosx}{\left(1+2sinx\right)\left(1-sinx\right)}=\sqrt{3}\)

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

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

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

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

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

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

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

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

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NL
23 tháng 9 2022 lúc 17:32

24.

ĐKXĐ: \(sin2x\ne0\Rightarrow x\ne\dfrac{k\pi}{2}\)

\(8sinx=\dfrac{1}{sinx}+\dfrac{\sqrt{3}}{cosx}\)

\(\Rightarrow8sinx.cosx.sinx=cosx+\sqrt{3}sinx\)

\(\Leftrightarrow4sin2x.sinx=sinx+\sqrt{3}cosx\)

\(\Leftrightarrow2cosx-2cos3x=cosx+\sqrt[]{3}sinx\)

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

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

\(\Leftrightarrow...\)

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