1d.

Đề ko rõ

1e.

\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)

\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)

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

\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)

\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)

\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)

2b.

Đề thiếu

2c.

Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)

\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)

\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)

\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)

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

\(\Leftrightarrow...\)

1.

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

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

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

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

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

Xét (1):

Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow...\)

2.

\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)

\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)

Xét (1):

Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm

@Nguyễn Việt Lâm giúp em với ạ

@Nguyễn Việt Lâm

7.

\(\Leftrightarrow\left[{}\begin{matrix}2x-40^0=60^0+k360^0\\2x-40^0=120^0+n360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=50^0+k180^0\\x=80^0+n180^0\end{matrix}\right.\)

Do \(-180^0\le x\le180^0\Rightarrow\left\{{}\begin{matrix}-180^0\le50^0+k180^0\le180^0\\-180^0\le80^0+n180^0\le180^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-\frac{23}{18}\le k\le\frac{13}{18}\\-\frac{13}{9}\le n\le\frac{5}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\left\{-1;0\right\}\\n=\left\{-1;0\right\}\end{matrix}\right.\)

\(\Rightarrow x=\left\{-130^0;50^0;-100^0;80^0\right\}\)

8.

\(\Leftrightarrow sinx=-\frac{\sqrt{2}}{2}\)

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

5.

\(\Leftrightarrow\frac{\sqrt{2}}{2}sin2x+\frac{\sqrt{2}}{2}cos2x=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin2x.sin\frac{\pi}{4}+cos2x.cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)

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

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

6.

\(\Leftrightarrow2sin2x=-1\)

\(\Leftrightarrow sin2x=-\frac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)