\(\Leftrightarrow1-2\sin^2x+2\sin x-1=0\)

\(\Leftrightarrow-2\sin^2x+2\sin x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin x=1\\\sin x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow2\cos^2x-1-3\cos x+2=0\)

\(\Leftrightarrow2\cos^2x-3\cos x+1=0\)

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

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

\(\Leftrightarrow1-\sin^2x-2\sin x+2=0\)

\(\Leftrightarrow-\sin^2x-2\sin x+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin x=1\\\sin x=-3\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)

\(y=f\left(x\right)=x^3-3x^2-9x+5\Rightarrow y'=3x^2-6x-9\)

giả sử d là tiếp tuyến của đths trên . Do d \(\perp\) đ/t : x - 9y +1 = 0

\(\Rightarrow y=-9x+k\) . Suy ra : \(3x^2-6x-9=-9\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

x = 0 \(\Rightarrow y=5\) .PTĐT d : \(y=-9x+5\)

\(x=2\Rightarrow y=-17\). PTĐT d : \(y=-9\left(x-2\right)-17=-9x+1\)

Chỗ y=-9x+k. K là sao vậy ạ

Câu 3 : 

Hàm số xác định khi sinx ≥ 0

⇔ \(k2\pi\le x\le\pi+k2\pi\) (nằm ở cung tròn phần dương của đường tròn lượng giác)

Câu 7

Hàm số xác định khi \(tan\left(x+\dfrac{\pi}{4}\right).tan\left(x-\dfrac{\pi}{4}\right)\ne0\)

⇔ \(\left\{{}\begin{matrix}tan\left(x-\dfrac{\pi}{4}\right)\ne0\\tan\left(x+\dfrac{\pi}{4}\right)\ne0\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x-\dfrac{\pi}{4}\ne k\pi\\x+\dfrac{\pi}{4}\ne k\pi\end{matrix}\right.\)⇔ \(x\ne\pm\dfrac{\pi}{4}+k\pi\) với k là số nguyên

Cảm ơn bạn nha

`\sqrt{3} sin 2x-cos 2x=2`

`<=>\sqrt{3}/2 sin2x-1/2cos 2x=1`

`<=>sin` `\pi/3 .sin 2x-cos` `\pi/3 .cos 2x=1`

`<=>cos(\pi/3 +2x)=-1`

`<=>\pi/3 + 2x=\pi+k2\pi` , `k in ZZ`

`<=>x=\pi/3+k\pi` , `k in ZZ`

\(Ta có:\) \(\sqrt{a^2+b^2}=\sqrt{\sqrt{3}^2+1^2}=2\)

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

\(\Leftrightarrow\sin2x\cdot\cos\dfrac{\pi}{6}-\cos2x\cdot\sin\dfrac{\pi}{6}=1\)

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

\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow2x=\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\) \(\left(k\in Z\right)\)

\(Ta\) \(có\): \(\sqrt{a^2+b^2}=\sqrt{\sqrt{3}^2+1^2}=2\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=\dfrac{1}{2}\)

\(\Leftrightarrow\sin x\cdot\cos\dfrac{\pi}{6}+\cos x\cdot\sin\dfrac{\pi}{6}=\dfrac{1}{2}\)

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

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

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

`\sqrt{3}sin x+cos x=1`

`<=>\sqrt{3}/2 sin x+1/2 cos x=1/2`

`<=>sin 60^o . sin x+cos 60^o . cos x=1/2`

`<=>cos(60^o -x)=1/2`

`<=>` $\left[\begin{matrix} 60^o -x=60^o +k360^o\\ 60^o-x=-60^o +k360^o\end{matrix}\right.$    `(k in ZZ)`

`<=>` $\left[\begin{matrix} x=-k360^o\\ x=120^2 -k360^o\end{matrix}\right.$    `(k in ZZ)`