\(sinx+cosx=m\Leftrightarrow\left(sinx+cosx\right)^2=m^2\)

\(\Leftrightarrow1+2sinx.cosx=m^2\Rightarrow sinx.cosx=\dfrac{m^2-1}{2}\)

\(A=sin^2x+cos^2x=1\)

\(B=sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)

\(=m^3-\dfrac{3m\left(m^2-1\right)}{2}=\dfrac{2m^3-3m^3+3m}{2}=\dfrac{3m-m^3}{2}\)

\(C=\left(sin^2+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1-2\left(\dfrac{m^2-1}{2}\right)^2\)

\(D=\left(sin^2x\right)^3+\left(cos^2x\right)^3=\left(sin^2x+cos^2x\right)^3-3\left(sin^2x+cos^2x\right)\left(sinx.cosx\right)^2\)

\(=1-3\left(\dfrac{m^2-1}{2}\right)^2\)

tham khảo:

(sinx + sin5x) + (sin2x + sin4x) + 4sin3x = 0

⇔ 2sin3x . cos2x + 2sin3x . cosx + 4sin3x = 0

⇔ 2sin3x (cos2x + cosx + 2sin3x) = 0

⇔ \(\left[{}\begin{matrix}sin3x=0\left(1\right)\\cos2x+cosx+2sin3x=0\left(2\right)\end{matrix}\right.\)

(1) ⇔ ...

(2) ⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}+4sin\dfrac{3x}{2}.cos\dfrac{3x}{2}=0\)

⇔ \(\left[{}\begin{matrix}cos\dfrac{3x}{2}=0\left(\alpha\right)\\cos\dfrac{x}{2}+2sin\dfrac{3x}{2}=0\left(\beta\right)\end{matrix}\right.\)

Giải \(\left(\alpha\right)\) quá đơn giản

Giải \(\left(\beta\right)\) 

\(2\left(3sin\dfrac{x}{2}-4sin^3\dfrac{x}{x}\right)+cos\dfrac{x}{2}=0\)

⇔ \(-8sin^3\dfrac{x}{2}+6sin\dfrac{x}{2}\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)+cos\dfrac{x}{2}.\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)=0\)

⇔ \(-2sin^3\dfrac{x}{2}+6sin\dfrac{x}{2}.cos^2\dfrac{x}{2}+sin^2\dfrac{x}{2}.cos\dfrac{x}{2}+cos^3\dfrac{x}{2}=0\) 

Xét \(x=k2\pi,k\in Z\) tức \(sin\dfrac{x}{2}=0\) có thỏa mãn phương trình không, nếu có kết luận về nghiệm 

Dù trường hợp trên có thỏa mãn hay không thì tiếp tục xét trường hợp nữa là \(x\ne k2\pi,k\in Z\) tức \(sin\dfrac{x}{2}\ne0\). Rồi chia cả 2 vế phương trình lằng nhằng kia cho \(sin\dfrac{x}{2}\) và đưa về phương trình bậc 3 theo cot\(\dfrac{x}{2}\)

Nếu tham khảo theo cách của mình thì dùng công thức này : 

sin3x

= sin2x . cosx + cos2x . sinx 

= 2sinx . cosx . cosx + (1 - 2sin²x) . sinx

= 2sinx . cos²x + sinx - 2sin³x

= 2sinx (1 - sin²x) + sinx - 2sin³x

= 3sinx - 4sin³x

1. Ta có: \(-1\le sinx\le1\)

\(\Rightarrow-3\le y\le3\) (hàm đã cho đồng biến trên \(\left[-\frac{\pi}{2};\frac{\pi}{2}\right]\)

\(y_{min}=-3\) khi \(sinx=-1\)

\(y_{max}=3\) khi \(sinx=1\)

2.

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

Do \(-1\le sinx\le1\Rightarrow0\le sinx+1\le2\)

\(\Rightarrow-2\le y\le2\)

\(y_{min}=-2\) khi \(sinx=1\)

\(y_{max}=2\) khi \(sinx=-1\)

3.

\(y=1-cos^2x+cos^4x=\left(cos^2x-\frac{1}{2}\right)^2+\frac{3}{4}\)

\(\Rightarrow y\ge\frac{3}{4}\Rightarrow y_{min}=\frac{3}{4}\) khi \(cos^2x=\frac{1}{2}\)

\(y=1+cos^2x\left(cos^2x-1\right)\le1\) do \(cos^2x-1\le0\)

\(\Rightarrow y_{max}=1\) khi \(\left[{}\begin{matrix}cos^2x=1\\cos^2x=0\end{matrix}\right.\)

4.

\(y=\left(sin^2x+cos^2x\right)^2-2\left(sinx.cosx\right)^2+sinx.cosx\)

\(y=1-\frac{1}{2}sin^22x+\frac{1}{2}sin2x\)

\(y=\frac{9}{8}-\frac{1}{2}\left(sinx-\frac{1}{2}\right)^2\le\frac{9}{8}\)

\(y_{max}=\frac{9}{8}\) khi \(sinx=\frac{1}{2}\)

\(y=\frac{1}{2}\left(sinx+1\right)\left(2-sinx\right)\ge0;\forall x\)

\(\Rightarrow y_{min}=0\) khi \(sinx=-1\)

a. ĐKXĐ: ...

\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)

b.

\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow4cos^32x-2cos2x-1=0\)

Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề

c. ĐKXĐ: ...

\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)

\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)

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

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

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

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

\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)

d.

\(\Leftrightarrow2\left(cosx-cos3x\right)sin4x=sin3x\)

\(\Leftrightarrow2sin4x.cosx-2sin4x.cos3x=sin3x\)

\(\Leftrightarrow sin5x+sin3x-sin7x-sinx=sin3x\)

\(\Leftrightarrow sin5x-sin7x-sinx=0\)

\(\Leftrightarrow-2cos6x.sinx-sinx=0\)

\(\Leftrightarrow sinx\left(2cos6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos6x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

a/ \(\Leftrightarrow2cosx.cos2x=cos2x\)

\(\Leftrightarrow2cosx.cos2x-cos2x=0\)

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

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

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

b/ \(\Leftrightarrow2sinx.sin2x=sinx\)

\(\Leftrightarrow2sinx.sin2x-sinx=0\)

\(\Leftrightarrow sinx\left(2sin2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\sin2x=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=\frac{\pi}{6}+k2\pi\\2x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)

c/ \(\Leftrightarrow sin3x-sinx+sin4x-sin2x=0\)

\(\Leftrightarrow2cos2x.sinx+2cos3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos2x+cos3x\right)=0\)

\(\Leftrightarrow2sinx.2cos\frac{5x}{2}.cos\frac{x}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{5x}{2}=0\\cos\frac{x}{2}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{5}+\frac{k4\pi}{5}\\x=\pi+k4\pi\end{matrix}\right.\)

d/ \(\Leftrightarrow sin3x-sinx-\left(sin4x-sin2x\right)=0\)

\(\Leftrightarrow2cos2x.sinx-2cos3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos2x-cos3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos2x=cos3x\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\2x=3x+k2\pi\\2x=-3x+k2\pi\end{matrix}\right.\)

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