1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)

\(\Leftrightarrow cos^2x+2cosx+1=0\)

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

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

\(\frac{sin^22x-4sin^2x}{sin^22x+4sin^2x-4}=\frac{4sin^2x.cos^2x-4sin^2x}{4sin^2x.cos^2x+4\left(sin^2x-1\right)}\)

\(=\frac{4sin^2x\left(cos^2x-1\right)}{4sin^2x.cos^2x-4cos^2x}=\frac{-4sin^4x}{4cos^2x\left(sin^2x-1\right)}=\frac{sin^4x}{cos^4x}=tan^4x\)

\(\frac{sin^22x+4sin^2x-4}{1-8sin^2x-cos4x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{1-8sin^2x-\left(1-2sin^22x\right)}=\frac{4sin^2x.cos^2x-4cos^2x}{2sin^22x-8sin^2x}\)

\(=\frac{-4cos^2x\left(1-sin^2x\right)}{8sin^2x.cos^2x-8sin^2x}=\frac{-4cos^2x.cos^2x}{-8sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{2sin^4x}=\frac{1}{2}cot^4x\)

\(\frac{cos2x}{cot^2x-tan^2x}=\frac{cos2x.sin^2x.cos^2x}{cos^4x-sin^4x}=\frac{\left(cos^2x-sin^2x\right).\left(2sinx.cosx\right)^2}{4\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)}=\frac{1}{4}sin^22x\)

a

\(\Leftrightarrow\left(3sinx-sin3x\right)cos3x+\left(3cosx+cos3x\right)sin3x+3\sqrt{3}cos4x=3\)

\(\Leftrightarrow\left(sinx.cos3x+sin3x.cosx\right)+\sqrt{3}cos4x=1\)

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

Tới đây thôi, mình lười ghi rồi =))

b

\(\Leftrightarrow\left(1-cos2x\right)\left(2sin^2x-1\right)\left(2sin^2+1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)

\(\Leftrightarrow\left(1-cos2x\right)\left(-cos2x\right)\left(2-cos2x\right)=cos2x\left(7cos^22x+3cos2x+4\right)\)

\(\Leftrightarrow-cos^22x+3cos2x-2=7cos^22x+3cos2x+4\)

\(\Leftrightarrow4cos^22x+3=0\)

=> pt vô nghiệm

\(4\left(cosx+1\right)+4\left(1-cos^2x\right)-5-3m=0\)

\(\Leftrightarrow-4cos^2x+4cosx+3=3m\)

Đặt \(f\left(x\right)=-4cos^2x+4cosx+3\)

\(f\left(x\right)=-\left(2cosx-1\right)^2+4\le4\)

\(f\left(x\right)=-4cos^2x+4cosx+8-5=4\left(cosx+1\right)\left(2-cosx\right)-5\ge-5\)

\(\Rightarrow-5\le f\left(x\right)\le4\)

\(\Rightarrow-5\le3m\le4\Rightarrow-\frac{5}{3}\le m\le\frac{4}{3}\)

\(\Leftrightarrow cos4x+cos2x-4sin^22x+1=0\)

\(\Leftrightarrow2cos^22x+1+cos2x-4\left(1-cos^22x\right)+1=0\)

\(\Leftrightarrow6cos^22x+cos2x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{1}{2}\\cos2x=-\frac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow...\)

Lời giải:

PT $\Leftrightarrow (\sin 2x-\cos 2x)(4\sin 2x+\cos 2x)=0$

$\Rightarrow \sin 2x=\cos 2x$ hoặc $4\sin 2x+\cos 2x=0$

Nếu $\sin 2x=\cos 2x$. Kết hợp với $\sin ^22x+\cos ^22x=1$ suy ra $\sin 2x=\cos 2x=\frac{\pm}{\sqrt{2}}$

$\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}$ với $k$ nguyên

Vì $x\in (0;\pi)$ nên $x=\frac{\pi}{8}$ hoặc $x=\frac{5\pi}{8}$

Nếu $4\sin 2x+\cos 2x=0$

$\Rightarrow \tan 2x=\frac{-1}{4}$

$\Rightarrow x=\frac{1}{2}k\pi +\frac{1}{2}\tan ^{-1}\frac{-1}{4}$

Vì $x\in (0;\pi)$ nên $x=\frac{1}{2}\pi +\frac{1}{2}\tan ^{-1}\frac{-1}{4};\pi +\frac{1}{2}\tan ^{-1}\frac{-1}{4}$

Vậy có $4$ nghiệm thỏa mãn.

Lời giải:

PT $\Leftrightarrow (\sin 2x-\cos 2x)(4\sin 2x+\cos 2x)=0$

$\Rightarrow \sin 2x=\cos 2x$ hoặc $4\sin 2x+\cos 2x=0$

Nếu $\sin 2x=\cos 2x$. Kết hợp với $\sin ^22x+\cos ^22x=1$ suy ra $\sin 2x=\cos 2x=\frac{\pm}{\sqrt{2}}$

$\Rightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}$ với $k$ nguyên

Vì $x\in (0;\pi)$ nên $x=\frac{\pi}{8}$ hoặc $x=\frac{5\pi}{8}$

Nếu $4\sin 2x+\cos 2x=0$

$\Rightarrow \tan 2x=\frac{-1}{4}$

$\Rightarrow x=\frac{1}{2}k\pi +\frac{1}{2}\tan ^{-1}\frac{-1}{4}$

Vì $x\in (0;\pi)$ nên $x=\frac{1}{2}\pi +\frac{1}{2}\tan ^{-1}\frac{-1}{4};\pi +\frac{1}{2}\tan ^{-1}\frac{-1}{4}$

Vậy có $4$ nghiệm thỏa mãn.

1. 4sin²x + 8cos²x-9=0

⇔ 4(sin²x+cos²x) + 4cos²x = 9

⇔ cos²x= \(\frac{9}{4}\)

⇔ cosx= \(\left[{}\begin{matrix}cosx=\frac{3}{2}\left(KTM\right)\\cosx=\frac{-3}{2}\left(KTM\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

2.

1-5sinx + 2cos²x=0

⇔1- 5sinx + 2(1-sin²x)=0

⇔ 2sin²x + 5sinx -3 =0

⇔\(\left[{}\begin{matrix}sinx=0,5\\sinx=-3\left(ktm\right)\end{matrix}\right.\)

Có sinx=0,5

⇔x=\(\left[{}\begin{matrix}x=\frac{\pi}{6}+2k\pi\\\frac{5\pi}{6}+2k\pi\end{matrix}\right.\left(k\in z\right)\)

Bạn sửa lại giúp mình câu 2 chỗ x đó là dấu ngoặc nhọn nhé, không phải dấu ngoặc vuông. Mình bị nhầm.