Chương 1: HÀM SỐ LƯỢNG GIÁC. PHƯƠNG TRÌNH LƯỢNG GIÁC

\(f'\left(x\right)=\left(sin^2x\right)'+4\cdot\left(sinx'\right)-5'\)

\(=2\cdot sinx\cdot cosx+4\cdot cosx=2cosx\left(sinx+2\right)\)

\(f'\left(x\right)=0\)

=>\(cosx\left(sinx+2\right)=0\)

=>\(cosx=0\)

=>\(x=\dfrac{\Omega}{2}+k\Omega\)

mà \(x\in\left[0;\dfrac{\Omega}{2}\right]\)

nên \(x=\dfrac{\Omega}{2}\)

\(f\left(\dfrac{\Omega}{2}\right)=sin^2\left(\dfrac{\Omega}{2}\right)+4\cdot sin\left(\dfrac{\Omega}{2}\right)-5\)

=1+4-5=0

\(f\left(0\right)=sin^20+4\cdot sin0-5=-5\)

=>Chọn D

Lời giải:
$A=\cos 2x-2\sin 5x\sin x=\cos 2x-2.\frac{-1}{2}[\cos (5x+x)-\cos (5x-x)]$

$=\cos 2x+\cos 6x-\cos 4x$

$=(\cos 2x+\cos 6x)-\cos 4x$

$=2\cos \frac{2x+6x}{2}\cos \frac{6x-2x}{2}-\cos 4x$

$=2\cos 4x\cos 2x-\cos 4x$

$=\cos 4x[2\cos 2x-1]$

Những đáp án A,B,C,D bạn đưa ra không có đáp án nào đúng cả.

\(sinx+cos\left(2x+\dfrac{\Omega}{3}\right)=0\)

=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=-sinx=sin\left(-x\right)\)

=>\(cos\left(2x+\dfrac{\Omega}{3}\right)=cos\left(\dfrac{\Omega}{2}+x\right)\)

=>\(\left[{}\begin{matrix}2x+\dfrac{\Omega}{3}=x+\dfrac{\Omega}{2}+k2\Omega\\2x+\dfrac{\Omega}{3}=-x-\dfrac{\Omega}{2}+k2\Omega\end{matrix}\right.\)

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

=>\(\left[{}\begin{matrix}x=\dfrac{5}{6}\Omega+k2\Omega\\x=-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\end{matrix}\right.\)

TH1: \(x=\dfrac{5}{6}\Omega+k2\Omega\)

\(0< =x< =2\Omega\)

=>\(0< =\dfrac{5}{6}\Omega+k2\Omega< =2\Omega\)

=>\(-\dfrac{5}{6}\Omega< =k2\Omega< =\dfrac{7}{6}\Omega\)

=>\(-\dfrac{5}{6}< =2k< =\dfrac{7}{6}\)

=>-5/12<=k<=7/12

mà k nguyên

nên k=0

TH2: \(x=-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\)

\(0< =x< =2\Omega\)

=>\(0< =-\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}< =2\Omega\)

=>\(\dfrac{5}{18}\Omega< =\dfrac{k2\Omega}{3}< =\dfrac{41}{18}\Omega\)

=>\(\dfrac{5}{18}< =\dfrac{2k}{3}< =\dfrac{41}{18}\)

=>\(\dfrac{5}{6}< =2k< =\dfrac{41}{6}\)

=>\(\dfrac{5}{12}< =k< =\dfrac{41}{12}\)

mà k nguyên

nên \(k\in\left\{1;2;3\right\}\)

=>Có 4 nghiệm thỏa mãn

PT\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{3}\right)=0\\sin\left(x-\dfrac{\pi}{3}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{3}\right)=sin0\\sin\left(x-\dfrac{\pi}{3}\right)=sin\text{​​}\text{​​}\dfrac{\pi}{2}\end{matrix}\right.\)

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

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

Vậy...

\(\pi< \alpha< \dfrac{3\pi}{2}\Rightarrow sin\alpha< 0\)\(\Rightarrow sin\alpha=-\sqrt{1-cos^2\alpha}=\dfrac{-\sqrt{15}}{4}\)

\(sin\left(\alpha-\dfrac{2\pi}{3}\right)=sin\alpha.cos\dfrac{2\pi}{3}-cos\alpha.sin\dfrac{2\pi}{3}\)

\(=\dfrac{-\sqrt{15}}{4}.\dfrac{-1}{2}-\dfrac{-1}{4}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{15}+\sqrt{3}}{8}\)

Đặt \(sin\left(x-\dfrac{\Omega}{3}\right)=a\left(a\in\left[-1;1\right]\right)\)

Phương trình sẽ trở thành \(a^2-a=0\)

=>a(a-1)=0

=>\(\left[{}\begin{matrix}a=0\\a=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\dfrac{\Omega}{3}\right)=0\\sin\left(x-\dfrac{\Omega}{3}\right)=1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x-\dfrac{\Omega}{3}=k\Omega\\x-\dfrac{\Omega}{3}=\dfrac{\Omega}{2}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{3}+k\Omega\\x=\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\)

\(\Omega< a< \dfrac{3}{2}\Omega\)

=>\(sina< 0\)

\(sin^2a+cos^2a=1\)

=>\(sin^2a=1-\dfrac{1}{16}=\dfrac{15}{16}\)

=>\(sina=-\dfrac{\sqrt{15}}{4}\)

\(sin\left(a-\dfrac{2}{3}\Omega\right)=sina\cdot cos\left(\dfrac{2}{3}\Omega\right)-cosa\cdot sin\left(\dfrac{2}{3}\Omega\right)\)

\(=-\dfrac{\sqrt{15}}{4}\cdot\dfrac{-1}{2}-\dfrac{-1}{4}\cdot\dfrac{\sqrt{3}}{2}\)

\(=\dfrac{\sqrt{15}+\sqrt{3}}{8}\)

Ba số \(2x-1;x;2x+1\) là một cấp số nhân khi:

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

\(\Leftrightarrow\left(2x\right)^2-1^2=x^2\)

\(\Leftrightarrow4x^2-1=x^2\)

\(\Leftrightarrow4x^2-x^2=1\)

\(\Leftrightarrow3x^2=1\)

\(\Leftrightarrow x^2=\dfrac{1}{3}\)

\(\Leftrightarrow x=\pm\sqrt{\dfrac{1}{3}}\)

\(\Leftrightarrow x=\pm\dfrac{\sqrt{3}}{3}\)

Vậy \(2x-1;x;2x+1\) là một cấp số nhân khi \(x=\pm\dfrac{\sqrt{3}}{3}\)

Để \(3;x-2;27\) là một cấp số nhân thì:

\(\left(x-2\right)^2=3\cdot27\)

\(\Leftrightarrow\left(x-2\right)^2=81\)

\(\Leftrightarrow\left(x-2\right)^2=9^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=9\\x-2=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-7\end{matrix}\right.\)

Vậy: \(3;x-2;27\) là cấp số nhân khi \(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-7\end{matrix}\right.\)

\(3.n^2=27\\ \Leftrightarrow n^2=\dfrac{27}{3}=9=3^2=\left(-3\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}n=3\\n=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-2=3.3\\x-2=3.\left(-3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=9\\x-2=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=11\\x=-7\end{matrix}\right.\)

Ta có hàm số: \(y=\dfrac{1-cosx}{tanx}\) hàm số được xác định khi:

\(\left\{{}\begin{matrix}cosx\ne0\\tanx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k\pi\\x\ne k\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

Tập xác định của y là:

\(D=R\backslash\left(\dfrac{k\pi}{2};k\in Z\right)\)