Question

[Nguyễn Linh Chi]

Giải phương trình lượng giác:

24)  \(\cos2x-\cos6x+4\left(3\sin x-4\sin^3x+1\right)=0\)

25) \(\sin^2x-2\sin x+2=\sin^23x\)

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Answer 1

[Nguyễn Linh Chi]

30. \(\tan x+\cot x=2\sin\left(x+\frac{\pi}{4}\right)\)

ĐK: \(x\ne\frac{k\pi}{2}\)

pt <=> \(\frac{1}{\sin x.\cos x}=2\sin\left(x+\frac{\pi}{4}\right)\)

<=> \(\frac{1}{\sin2x}=\sin\left(x+\frac{\pi}{4}\right)\)

Đánh giá: \(-1\le\sin2x\le1\)

=> \(\orbr{\begin{cases}\frac{1}{\sin2x}\le-1\\\frac{1}{\sin2x}\ge1\end{cases}}\)

\(-1\le\sin\left(x+\frac{\pi}{4}\right)\le1\)

Như vậy dấu "=" xảy ra <=> \(\orbr{\begin{cases}\frac{1}{\sin2x}=\sin\left(x+\frac{\pi}{4}\right)=-1\\\frac{1}{\sin2x}=\sin\left(x+\frac{\pi}{4}\right)=1\end{cases}}\)

<=> \(\orbr{\begin{cases}\sin2x=\sin\left(x+\frac{\pi}{4}\right)=-1\\\sin2x=\sin\left(x+\frac{\pi}{4}\right)=1\end{cases}}\)

TH1: \(\sin2x=\sin\left(x+\frac{\pi}{4}\right)=-1\)

<=> \(\hept{\begin{cases}2x=-\frac{\pi}{2}+k2\pi\\x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{\pi}{4}+k\pi\\x=-\frac{3\pi}{4}+k2\pi\end{cases}}\)loại

TH2: 

 \(\sin2x=\sin\left(x+\frac{\pi}{4}\right)=1\)

<=> \(\hept{\begin{cases}2x=\frac{\pi}{2}+k2\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{4}+k2\pi\end{cases}}\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)

Vậy ...

Answer 2

[Nguyễn Linh Chi]

29) \(\sin x-2\sin2x-\sin3x=2\sqrt{2}\)

<=> \(\left(\sin x-\sin3x\right)-2\sin2x=2\sqrt{2}\)

<=> \(-2.\sin x\cos2x-2\sin2x=2\sqrt{2}\)

<=> \(\sin x\cos2x+\sin2x=-\sqrt{2}\)

Ta có: \(\left(\sin x\cos2x+\sin2x\right)^2\le\left(\sin^2x+1\right)\left(\sin^22x+\cos^22x\right)=\sin^2x+1\le2\)

( theo bunhia)

=> \(-\sqrt{2}\le\sin x\cos2x+\sin2x\le\sqrt{2}\)

Dấu "=" xảy ra <=> \(\frac{\sin x}{1}=\frac{\cos2x}{\sin2x}\)(1) và \(\sin x\cos2x+\sin2x=-\sqrt{2}\)(2)

(1) <=> \(\frac{\sin x.\cos2x}{1}=\frac{\cos^22x}{\sin2x}\)=> (2) <=>  \(\frac{\cos^22x}{\sin2x}+\sin2x=-\sqrt{2}\)

<=> \(\frac{1}{\sin2x}=-\sqrt{2}\)<=> \(\sin2x=-\frac{\sqrt{2}}{2}\)<=> \(\orbr{\begin{cases}x=-\frac{\pi}{8}+k\pi\\x=-\frac{3\pi}{8}+k\pi\end{cases}}\)

(1) <=> \(\sin x.\sin2x=\cos2x\)=> (2) <=> \(\sin x.\sin x.\sin2x+\sin2x=-\sqrt{2}\)

<=> \(\frac{\sin^2x}{2}+\frac{1}{2}=+1\Leftrightarrow\sin^2x=1\)=> \(\cos^2x=0\)loại vì \(\sin2x=-\frac{\sqrt{2}}{2}\)

Vậy pt vô nghiệm