Bài 1: Hàm số lượng giác

Đặt \(sinx=t\in\left[-1;1\right]\)

\(y=f\left(t\right)=t^2+2t\)

Xét hàm \(y=f\left(t\right)=t^2+2t\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=-1\in\left[-1;1\right]\)

\(f\left(-1\right)=-1\) ; \(f\left(1\right)=3\)

\(\Rightarrow y_{min}=-1\) khi \(sinx=-1\Rightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=3\) khi \(sinx=1\Rightarrow x=\dfrac{\pi}{2}+k2\pi\)

ĐKXĐ:

a. \(x-1\ne0\Rightarrow x\ne1\)

b. \(sinx+1>0\Rightarrow sinx\ne-1\Rightarrow x\ne-\dfrac{\pi}{2}+k2\pi\)

c. \(cos\left(2x-\dfrac{\pi}{6}\right)\ne0\Rightarrow2x-\dfrac{\pi}{6}\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{\pi}{3}+\dfrac{k\pi}{2}\)

d. \(cos\left(3x-\pi\right)\ne0\Rightarrow cos3x\ne0\Rightarrow3x\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)

e. \(sin^2x-cos^2x\ne0\Rightarrow cos2x\ne0\Rightarrow2x\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

g. \(cos2x-1\ne0\Rightarrow cos2x\ne1\Rightarrow2x\ne k2\pi\Rightarrow x\ne k\pi\)

a.

\(-1\le sin\left(x+\dfrac{\pi}{4}\right)\le1\Rightarrow-1\le y\le3\)

\(y_{min}=-1\) khi \(sin\left(x+\dfrac{\pi}{4}\right)=1\Rightarrow x=\dfrac{\pi}{4}+k2\pi\)

\(y_{max}=3\) khi \(sin\left(x+\dfrac{\pi}{4}\right)=-1\Rightarrow x=-\dfrac{3\pi}{4}+k2\pi\)

b.

\(0\le cos^23x\le1\Rightarrow2\le y\le3\)

\(y_{min}=2\) khi \(cos^23x=1\Rightarrow x=\dfrac{k\pi}{3}\)

\(y_{max}=3\) khi \(cos^23x=0\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)

c.

\(y=1-sin^2x+sinx+2=-sin^2x+sinx+3\)

Đặt \(sinx=t\in\left[-1;1\right]\) \(\Rightarrow y=f\left(t\right)=-t^2+t+3\)

\(-\dfrac{b}{2a}=\dfrac{1}{2}\in\left[-1;1\right]\)

\(f\left(-1\right)=1\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{13}{4}\); \(f\left(1\right)=3\)

\(\Rightarrow y_{min}=1\) khi \(sinx=-1\Rightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=\dfrac{13}{4}\) khi \(sinx=\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Hàm số xác định khi:

\(\left\{{}\begin{matrix}cos\dfrac{x}{2}\ne0\\cosx+1\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}\ne\dfrac{\pi}{2}+k\pi\\cosx\ne-1\end{matrix}\right.\)

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

1.

\(2cos^2x-3cosx+1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

7.

\(cos2x-5sinx-3=0\)

\(\Leftrightarrow1-2sin^2x-5sinx-3=0\)

\(\Leftrightarrow2sin^2x+5sinx+2=0\)

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

\(\Leftrightarrow sinx=-\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)

8.

ĐK: \(x\ne\dfrac{k\pi}{2}\)

\(5tanx-2cotx-3=0\)

\(\Leftrightarrow5tanx-\dfrac{2}{tanx}-3=0\)

\(\Leftrightarrow5tan^2x-3tanx-2=0\)

\(\Leftrightarrow\left(tanx-1\right)\left(5tanx+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=-\dfrac{2}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(-\dfrac{2}{5}\right)+k\pi\end{matrix}\right.\)

\(-1\le sin\left(x^2\right)\le1\Rightarrow\)\(0\le\sqrt{1-sin\left(x^2\right)}\le\sqrt{2}\Rightarrow-1\le y\le\sqrt{2}-1\)

\(y_{min}=-1\) khi \(sin\left(x^2\right)=1\Rightarrow x=\pm\sqrt{\dfrac{\pi}{2}+k2\pi}\) (\(k\in N\))

\(y_{max}=\sqrt{2}-1\) khi \(sin\left(x^2\right)=-1\Rightarrow x=\pm\sqrt{-\dfrac{\pi}{2}+k2\pi}\) (\(k\in Z^+\))

Đề bài yêu cầu điều gì nhỉ?

\(\Leftrightarrow\left(\sqrt{3}+2\right)sinx+cosx=2sin3x+2sinx\)

\(\Leftrightarrow\sqrt{3}sinx+cosx=2sin3x\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=sin3x\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=sin3x\)

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

\(\Leftrightarrow x=...\)

\(0\le sin^23x\le1\Rightarrow3\le y\le4\)

\(y_{min}=3\) khi \(sin^23x=1\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)

\(y_{max}=4\) khi \(sin^23x=0\Rightarrow x=\dfrac{k\pi}{3}\)