Toán tui ngu lắm, mấy tháng rồi chưa sờ đến toán luôn :) Nhờ tui làm chi

a/ \(2\cos^24x=1+\cos x.\cos3x-\sin x.\sin3x\)

\(\Leftrightarrow2\cos^24x=1+\cos\left(3x+x\right)=1+\cos4x\)

\(\Leftrightarrow2\cos^24x-\cos4x-1=0\)

Pt bậc 2, bạn tự giải

b/ \(\Leftrightarrow2.\frac{1}{2}\left[\cos2x-\cos8x\right]+\cos4x=\cos2x\)

\(\cos8x=\cos2.4x=2\cos^24x-1\)

\(\Rightarrow-2\cos^24x+1+\cos4x=0\)

\(\Leftrightarrow2\cos^24x-\cos4x-1=0\)

Pt bậc 2, bạn tự giải

\(sin^2x+cosx.cos3x+sin2x.cos2x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}cos4x+\dfrac{1}{2}cos2x+\dfrac{1}{2}sin4x=0\)

\(\Leftrightarrow sin^2x+\dfrac{1}{2}-sin^2x+\dfrac{1}{2}sin4x+\dfrac{1}{2}cos4x=0\)

\(\Leftrightarrow sin4x+cos4x=-1\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Lớp 11 thì Bó Tay 

\(x=10\)

mk ấn máy tính thì nó bằng thế

mk mới lp 9 thôi mà

\(sin4x=-cos2x\\ \Leftrightarrow sin4x+cos2x=0\\ \Leftrightarrow2sin2x.cos2x+cos2x=0\\ \Leftrightarrow cos2x\left(2sin2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}cos2x=0\\2sin2x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{\pi}{2}+k\pi\\sin2x=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\sin2x=sin\left(-\dfrac{\pi}{6}\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{matrix}\right.\)

`HaNa♫D`

a.

\(\Leftrightarrow cosx\left[1-\left(1-2sin^2x\right)\right]-sin^2x=0\)

\(\Leftrightarrow2sin^2x.cosx-sin^2x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\frac{1}{2}\end{matrix}\right.\)

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

b.

Câu b chắc chắn đề đúng chứ bạn? Vế phải ấy?

c/

\(\left(1+cosx\right)\left(sinx-cosx+3\right)=1-cos^2x\)

\(\Leftrightarrow\left(1+cosx\right)\left(sinx-cosx+3\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)

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

\(\Leftrightarrow cosx=-1\)

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

d.

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

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

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

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

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

tanx = 1-cos2x (ĐK x\(\ne\dfrac{\pi}{2}+k\pi\))

\(\Leftrightarrow\dfrac{sinx}{cosx}=2sin^2x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sin2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

P/t \(\Leftrightarrow2cos2x.sin2x-sin2x+2cos^22x-cos2x-1=0\)

\(\Leftrightarrow sin4x-sin2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sinx.cos3x-2sin3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(1\right)\\cos3x=sin3x\left(2\right)\end{matrix}\right.\) 

(1) \(\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) \(\Leftrightarrow sin3x-cos3x=0\)  \(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\left(k\in Z\right)\)

Vậy ... 

1.

\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)

\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)

\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)

\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)

Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)

\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm:

\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)

2.

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

\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)

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

\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)

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

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)