a/

\(\Leftrightarrow sinx.cosx\left(sin^2x-cos^2x\right)=\frac{\sqrt{2}}{8}\)

\(\Leftrightarrow2sinx.cosx\left(cos^2x-sin^2x\right)=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow sin2x.cos2x=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow\frac{1}{2}sin4x=-\frac{\sqrt{2}}{4}\)

\(\Leftrightarrow sin4x=-\frac{\sqrt{2}}{2}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{16}+\frac{k\pi}{2}\\x=\frac{5\pi}{16}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

Câu này đề hơi kì quái, bạn coi lại đề được ko? Biến đổi mấy cách vẫn thấy ko ổn

c/

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

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

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

\(\left(1\right)\Leftrightarrow cosx=-1\Leftrightarrow\pi x=\pi+k2\pi\)

\(\left(2\right)\Leftrightarrow2sinx=0\Rightarrow sinx=0\)

\(\Rightarrow x=k\pi\)

Kết hợp lại ta được \(x=k\pi\)

d/

\(\Leftrightarrow2sin8x.cosx=cos\left(\frac{\pi}{2}-2x\right)+1-1-cos\left(\frac{\pi}{2}+4x\right)\) (hạ bậc vế phải)

\(\Leftrightarrow2sin8x.cosx=sin2x+sin4x\)

\(\Leftrightarrow2sin8x.cosx=2sin3x.cosx\)

\(\Leftrightarrow cosx\left(sin8x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin8x=sin3x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=3x+k2\pi\\8x=\pi-3x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{k2\pi}{5}\\x=\frac{\pi}{11}+\frac{k2\pi}{11}\end{matrix}\right.\)

a)\(pt\Leftrightarrow\frac{1-cos8x}{2}+\frac{1-cos6x}{2}=\frac{1-cos4x}{2}+\frac{1-cos2x}{2}\)

\(\Leftrightarrow cos2x+cos4x=cos6x+cos8x\)

\(\Leftrightarrow2cos3x\cdot cosx=2cos7x\cdot cosx\)

\(\Leftrightarrow2cos\left(cos3x-cos7x\right)=0\)

\(\Leftrightarrow2cosx\cdot\left(-2\right)\cdot sin5x\cdot sin\left(-2x\right)=0\)

\(\Leftrightarrow cosx\cdot sin2x\cdot sin5x=0\)

\(\Leftrightarrow sin2x\cdot sin5x=0\)(do sin2x=0 <=>2sinx*cosx=0 gồm th cosx=0 r`)

\(\Leftrightarrow\left[\begin{array}{nghiempt}sin2x=0\\sin5x=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{k\pi}{2}\\x=\frac{k\pi}{5}\end{array}\right.\)\(\left(k\in Z\right)\)

b)\(pt\Leftrightarrow1-cos2x+1-cos4x=1+cos6x+1+cos8x\)

\(\Leftrightarrow cos2x+cos8x+cos4x+cos6x=0\)

\(\Leftrightarrow cos10x\cdot cos6x+cos10x\cdot cos2x=0\)

\(\Leftrightarrow cos10x\left(cos6x+cos2x\right)=0\)

\(\Leftrightarrow cos10x\cdot cos8x\cdot cos4x=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}cos10x=0\\cos8x=0\\cos4x=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{20}+\frac{k\pi}{10}\\x=\frac{\pi}{16}+\frac{k\pi}{8}\\x=\frac{\pi}{8}+\frac{k\pi}{4}\end{array}\right.\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x=\frac{1}{2}+\frac{1}{2}cos4x+\frac{1}{2}+\frac{1}{2}cos8x\)

\(\Leftrightarrow cos8x+cos2x+cos6x+cos4x=0\)

\(\Leftrightarrow2cos5x.cos3x+2cos5x.cosx=0\)

\(\Leftrightarrow cos5x\left(cos3x+cosx\right)=0\)

\(\Leftrightarrow2cos5x.cos2x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos5x=0\\cos2x=0\\cosx=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{10}+\frac{k\pi}{5}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left(sin4x-cos4x\right)\left(sin4x+4cos4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x-cos4x=0\\sin4x=-4cos4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(4x-\frac{\pi}{4}\right)=0\\tan4x=-4\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{16}+\frac{k\pi}{4}\\x=\frac{1}{4}arctan\left(-4\right)+\frac{k\pi}{4}\end{matrix}\right.\)

Pt có 4 nghiệm trên khoảng đã cho

a/

\(\Leftrightarrow\left(sin^2\frac{x}{3}+cos^2\frac{x}{3}\right)^2-2sin^2\frac{x}{3}.cos^2\frac{x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{2}sin^2\frac{2x}{3}=\frac{5}{8}\)

\(\Leftrightarrow1-\frac{1}{4}\left(1-cos\frac{4x}{3}\right)=\frac{5}{8}\)

\(\Leftrightarrow cos\frac{4x}{3}=-\frac{1}{2}\)

\(\Leftrightarrow\frac{4x}{3}=\pm\frac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\frac{\pi}{2}+\frac{k3\pi}{2}\)

b/

\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

\(\Leftrightarrow4-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)

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

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

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

\(\Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=-\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

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

c/

\(\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)

\(\Leftrightarrow-cos^32x+5cos^22x-7cos2x+3=0\)

\(\Leftrightarrow\left(3-cos2x\right)\left(cos2x-1\right)^2=0\)

\(\Leftrightarrow cos2x=1\)

\(\Leftrightarrow x=k\pi\)

d/

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos4x\)

\(\Leftrightarrow1-\frac{3}{4}sin^22x=cos4x\)

\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)=cos4x\)

\(\Leftrightarrow cos4x=1\)

\(\Leftrightarrow x=\frac{k\pi}{2}\)

1.

\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow x=...\)

(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)

2.

\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)

\(\Leftrightarrow2cos^22x-cos2x=cos2x\)

\(\Leftrightarrow cos^22x-cos2x=0\)

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

3.

\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)

\(\Leftrightarrow...\)

4.

\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)

\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)

\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)

\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)

\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)

\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)

5.

\(cos^22x+\frac{1}{2}+\frac{1}{2}cos6x=\frac{1}{2}-\frac{1}{2}cos2x\)

\(\Leftrightarrow cos^22x+\frac{1}{2}\left(cos6x+cos2x\right)=0\)

\(\Leftrightarrow cos^22x+cos4x.cos2x=0\)

\(\Leftrightarrow cos2x\left(cos2x+cos4x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-1\\cos2x=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

Quên cách giải ptlg rồi nên lm câu 4 =.=

\(\cos3x=\cos\left(2x+x\right)=\cos2x.\cos x-\sin2x.\sin x\)

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

\(=2\cos^3x-\cos x-2\sin^2x.\cos x\)

\(\Rightarrow A=\frac{1+\cos x+2\cos^2x-1+2\cos^3x-\cos x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\left(1-\cos^2x\right).\cos x}{2\cos^2x-1+\cos x}\)

\(=\frac{2\cos^2x+2\cos^3x-2\cos x+2\cos^3x}{2\cos^2x-1+\cos x}\)

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

a/ \(cosx-cos2x+sin2x-sinx=3-4cosx\)

\(\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)

b/ ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.\) \(\Rightarrow...\)

\(\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}\)

\(\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)

\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)

\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)

\(A=0\)

\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)

\(B=\frac{1}{4}\)

\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)

\(C=\frac{3}{2}\)

\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)

\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)

\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)

\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)

Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là