a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)

Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).

\(\Rightarrow y=f\left(t\right)=3-t^2\)

\(\Rightarrow y_{min}=minf\left(t\right)=2\)

\(y_{max}=maxf\left(t\right)=3\)

b, \(y=f\left(t\right)=\dfrac{-2}{3t-5}\left(t\in\left[0;1\right]\right)\)

\(\Rightarrow y_{min}=minf\left(t\right)=\dfrac{2}{5}\)

\(y_{max}=maxf\left(t\right)=1\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3\left(1-cos^2x\right)-4cos^2x}{cos^2x}dx=\int\limits^{\dfrac{\pi}{4}}_0\dfrac{3-7cos^2x}{cos^2x}dx\)

\(=\int\limits^{\dfrac{\pi}{4}}_0\left(\dfrac{3}{cos^2x}-7\right)dx=\left(3tanx-7x\right)|^{\dfrac{\pi}{4}}_0=...\)

b: \(cosx+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(cos\left(2\cdot\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(1-2\cdot\sin^2\left(\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-2=0\)

=>\(-2\cdot\sin^2\left(\frac{x}{2}\right)+3\cdot\sin\left(\frac{x}{2}\right)-1=0\)

=>\(2\cdot\sin^2\left(\frac{x}{2}\right)-3\cdot\sin\left(\frac{x}{2}\right)+1=0\)

=>\(\left(2\cdot\sin\left(\frac{x}{2}\right)-1\right)\left(\sin\left(\frac{x}{2}\right)-1\right)=0\)

TH1: \(\sin\left(\frac{x}{2}\right)-1=0\)

=>\(\sin\left(\frac{x}{2}\right)=1\)

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

=>\(x=\pi+k4\pi\)

TH2: \(2\cdot\sin\left(\frac{x}{2}\right)-1=0\)

=>\(\sin\left(\frac{x}{2}\right)=\frac12\)

=>\(\left[\begin{array}{l}\frac{x}{2}=\frac{\pi}{6}+k2\pi\\ \frac{x}{2}=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k4\pi\\ x=\frac53\pi+k4\pi\end{array}\right.\)

A

Haha, anh nhanh thía :P

d.

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^4x\)

\(tan^4x-3tan^2x-4tanx-3=0\)

\(\Leftrightarrow\left(tan^2x+tanx+1\right)\left(tan^2x-tanx-3\right)=0\)

\(\Leftrightarrow tan^2x-tanx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1-\sqrt{13}}{2}\\tanx=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\frac{1-\sqrt{13}}{2}\right)+k\pi\\x=arctan\left(\frac{1+\sqrt{13}}{2}\right)+k\pi\end{matrix}\right.\)

mọi người giúp hộ mình nhanh với

Đáp án C