a/

Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)

Pt trở thành:

\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)

\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)

\(\Leftrightarrow cos^2a+4sina-4=0\)

\(\Leftrightarrow1-sin^2a+4sina-4=0\)

\(\Leftrightarrow-sin^2a+4sina-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)

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

\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)

b/

Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)

Pt trở thành:

\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)

\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)

\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)

\(\Leftrightarrow10sin^2a+4sina-14=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)

\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

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

c/

\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)

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

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

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=...\)

1.

\(0\le cos^2\left(\frac{x}{2}-\frac{\pi}{9}\right)\le1\)

\(\Rightarrow-5\le y\le-1\)

\(y_{min}=-5\) khi \(cos\left(\frac{x}{2}-\frac{\pi}{9}\right)=0\)

\(y_{max}=-1\) khi \(cos^2\left(\frac{x}{2}-\frac{\pi}{9}\right)=1\)

2.

Hàm \(y=3-sin7x\) có chu kì \(T=\frac{2\pi}{7}\)

Hàm \(y=\frac{sin2x.cos2x}{25}=\frac{1}{50}sin4x\) có chu kì \(T=\frac{2\pi}{4}=\frac{\pi}{2}\)

Tại sao lại suy ra được -5≤y≤-1 vậy

x.l nha mik ms hkl p 7 àk

Google có nha bn

\(cos2\left(x+\frac{\pi}{6}\right)+4cos\left(\frac{\pi}{3}-x\right)=\frac{5}{2}\)

\(4sin\left(x+\frac{\pi}{6}\right)+\left(x+\frac{\pi}{6}\right)cos2=\frac{5}{2}\)

\(\frac{1}{6}\left(24sin\right)\left(x+\frac{\pi}{6}\right)+6x\left(cos2\right)=\frac{5}{2}\)

\(2\sqrt{3}sin\left(x\right)+x\)\(cos\left(2\right)+2cos\left(x\right)+\frac{1}{6}\pi\)\(cos\left(2\right)=\frac{5}{2}\)

\(\left(2\sqrt[6]{-1}-2\left(-1^{\frac{5}{6}}\right)\right)sin\left(x\right)+x\left(cos2\right)+\left(2\sqrt[3]{-1-2\left(-1^{\frac{2}{3}}\right)}\right)cos\left(x\right)=\frac{5}{2}-\frac{1}{6}\pi\)\(cos\left(2\right)\)

\(24sin\left(x+\frac{\pi}{6}\right)+\left(6x+\pi\right)cos\left(2\right)=15\)

\(4sin\left(x+\frac{\pi}{6}\right)+x\)\(cos\left(2\right)+\frac{1}{6}\pi\)\(cos\left(2\right)=\frac{5}{2}\)

\(\Rightarrow x=\left\{-15,1252;-13,976;-6,8388;-3,93832\right\}\)

\(cos^2\left(x+\frac{\pi}{6}\right)\) hả

uk đúng rồi ...mk viết nhầm ..phải là cos^2 nha

èo,,,nhìn khó tek,,,,lm đi má ơi

a/

\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)

\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)

\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)

\(\Leftrightarrow-sin4x=sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)

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

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

b/

\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)

\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)

\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow1-sin^2\left(x+\frac{\pi}{3}\right)+4sin\left(x+\frac{\pi}{3}\right)-4=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)-4sin\left(x+\frac{\pi}{3}\right)+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=1\\sin\left(x+\frac{\pi}{3}\right)=3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow...\)