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

\(\Leftrightarrow\left(2cosx-1\right)\left(2cosx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cosx=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow x=...\)

1) \(2sin^2x-5sinx-3=0\Rightarrow\left\{{}\begin{matrix}sinx=3\left(l\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

    \(\Rightarrow sinx=sin\left(-\dfrac{\pi}{6}\right)\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (k\(\in\)Z)

2) \(\sqrt{3}cot^2x-4cotx+\sqrt{3}=0\)

    Đk: \(x\ne m\pi\)

    Pt; \(\Rightarrow\left[{}\begin{matrix}cotx=\sqrt{3}\\cotx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\) bạn tự tìm x nhé

\(\Leftrightarrow\left(sinx-3\right)\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=3>1\left(loại\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow sin8x-\sqrt{2}cos8x=cos6x-\sqrt{2}sin6x\)

\(\Leftrightarrow\dfrac{1}{\sqrt{3}}sin8x-\dfrac{\sqrt{2}}{\sqrt{3}}cos8x=\dfrac{1}{\sqrt{3}}cos6x-\dfrac{\sqrt{2}}{\sqrt{3}}sin6x\)

Đặt \(\dfrac{1}{\sqrt{3}}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{\sqrt{2}}{\sqrt{3}}=sina\)

\(\Rightarrow sin8x.cosa-cos8x.sina=cos6x.cosa-sin6x.sina\)

\(\Leftrightarrow sin\left(8x-a\right)=cos\left(6x+a\right)\)

\(\Leftrightarrow sin\left(8x-a\right)=sin\left(\dfrac{\pi}{2}-6x-a\right)\)

\(\Leftrightarrow...\)

a) \(y=1-2sinx\)

Ta có: \(-1\le sinx\le1\Rightarrow-2\le2sinx\le2\)

                                   \(\Rightarrow2\ge-2sin2x\ge-2\)

                                   \(\Rightarrow3\ge1-2sinx\ge-1\)

      Vậy \(y_{max}=3,y_{min}=-1\)

1.C

\(sinx+2=0\Rightarrow sinx=-2< -1\) vô nghiệm

2.D

\(tan5x=-\dfrac{9}{5}\) phương trình \(tanu=a\) luôn luôn có nghiệm

3B

\(y=cosx-cos\left(x-\dfrac{2\pi}{3}\right)+3=-2sin\left(x-\dfrac{\pi}{3}\right).sin\left(\dfrac{\pi}{3}\right)+3\)

\(=-\sqrt{3}sin\left(x-\dfrac{\pi}{3}\right)+3\ge3-\sqrt{3}\)

4D

\(cos3x=-\dfrac{\sqrt{3}}{2}\Leftrightarrow3x=\pm\dfrac{5\pi}{6}+k2\pi\)

\(\Rightarrow x=\pm\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\) có 3 nghiệm trên khoảng đã cho \(x=\left\{\dfrac{5\pi}{18};\dfrac{7\pi}{18};\dfrac{17\pi}{18}\right\}\)

5.

\(cos\left(5x-1\right)=m^2\)

Do \(-1\le cos\left(5x-1\right)\le1\Rightarrow-1\le m^2\le1\)

\(\Rightarrow m^2\le1\)

\(\Rightarrow-1\le m\le1\)

6.

\(5sinx-m+2=0\Rightarrow sinx=\dfrac{m-2}{5}\)

Do \(-1\le sinx\le1\) nên pt có nghiệm khi:

\(-1\le\dfrac{m-2}{5}\le1\Rightarrow-3\le m\le7\)

7.

\(\left(sin2x-cosx\right)\left(2cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=cosx\\2cosx=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=sin\left(\dfrac{\pi}{2}-x\right)\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{2}+k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

1.

\(\Leftrightarrow2cos2x+sinx-sin3x=0\)

\(\Leftrightarrow2cos2x-2cos2x.sinx=0\)

\(\Leftrightarrow2cos2x\left(1-sinx\right)=0\)

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

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

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

2.

\(cos^2x+\left(sin3x-1\right)\left(1-cos\left(\dfrac{\pi}{2}-x\right)\right)=0\)

\(\Leftrightarrow1-sin^2x+\left(sin3x-1\right)\left(1-sinx\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx\right)+\left(sin3x-1\right)\left(1-sinx\right)=0\)

\(\Leftrightarrow\left(1-sinx\right)\left(1+sinx+sin3x-1\right)=0\)

\(\Leftrightarrow2\left(1-sinx\right)sin2x.cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sin2x=0\\cosx=0\end{matrix}\right.\)

\(\Leftrightarrow sin2x=0\)

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