1) cot2x=-√3/3
2)
a) tan^2x+(1-√3)tanx-√3=0
b) 4sin^2x-4cosx-1=0
c)1/cos^2x -(3+√3)tanx-1+√3=0
1) cot2x=-√3/3 2) a) tan^2x+(1-√3)tanx-√3=0 b) 4sin^2x-4cosx-1=0 c)1/cos^2x -(3+√3)tanx-1+√3=0
By Allison
By Allison
1) cot2x=-√3/3
2)
a) tan^2x+(1-√3)tanx-√3=0
b) 4sin^2x-4cosx-1=0
c)1/cos^2x -(3+√3)tanx-1+√3=0
$1/cot2x=\dfrac{-\sqrt{3}}{3}\\\Leftrightarrow\left[\begin{matrix}2x=\dfrac{2π}{3}+kπ\\2x=\dfrac{-π}{3}+kπ\end{matrix}\right.(k\in Z)$
$\Leftrightarrow\left[\begin{matrix}x=\dfrac{π}{3}+kπ\\x=\dfrac{-π}{6}+kπ\end{matrix}\right.(k\in Z)$
$a)\ tan^2x+(1-\sqrt{3})tanx-\sqrt{3}=0\\\Leftrightarrow \left[\begin{matrix}tanx=-1\\tanx=\sqrt{3}\end{matrix}\right.$
$*)tanx=-1\\\Leftrightarrow\left[\begin{matrix}x=\dfrac{3π}{4}+kπ\\x=\dfrac{-π}{4}+kπ\end{matrix}\right.(k\in Z)$
$*) tanx=\sqrt{3}\\\Leftrightarrow\left[\begin{matrix}x=\dfrac{π}{3}+kπ\\x=\dfrac{-2π}{3}+kπ\end{matrix}\right.(k\in Z)$
$b)\ 4sin^2x-4cosx-1=0\\\Leftrightarrow 4sin^2x-4cosx-4+3=0\\\Leftrightarrow 4sin^2x-4cosx-4sin^2x-4cos^2x+3=0\\\Leftrightarrow 4cos^2x+4cosx-3=0\\\Leftrightarrow \left[\begin{matrix}cosx=\dfrac{-3}{2}\ (loại)\\cosx=\dfrac{1}{2}\ (t/m)\end{matrix}\right.$
$cosx=\dfrac{1}{2}\\\Leftrightarrow \left[\begin{matrix}x=\dfrac{π}{3}+k2π\\x=\dfrac{-π}{3}+k2π\end{matrix}\right.(k\in Z)$
$c)\ \dfrac{1}{cos^2x}-(3+\sqrt{3})tanx-1+\sqrt{3}=0\\\Leftrightarrow tan^2x-(3+\sqrt{3})tanx+\sqrt{3}=0.$
Câu này hình như đề sai phải là
$\dfrac{1}{cos^2x}-(1+\sqrt{3})tanx-1+\sqrt{3}=0$
đúng không bạn
Đáp án:
Giải thích các bước giải:
1) `cot\ 2x=-\frac{\sqrt{3}}{3}`
ĐK: `sin\ 2x \ne 0`
`⇔ 2x \ne k\pi\ (k \in \mathbb{Z})`
`⇔ x \ne k\frac{\pi}{2}\ (k \in \mathbb{Z})`
`⇒ cot 2x=cot (\frac{2\pi}{3})`
`⇔ 2x=\frac{2\pi}{3}+k\pi\ (k \in \mathbb{Z})`
`⇔ x=\frac{\pi}{3}+k\frac{\pi}{2}\ (k \in \mathbb{Z})` (TM)
Vậy `S={\frac{\pi}{3}+k\frac{\pi}{2}\ (k \in \mathbb{Z})}`
2)
a) `tan^2 x+(1-\sqrt{3})tan\ x-\sqrt{3}=0`
`⇔ (tan\ x+1)(tan\ x-\sqrt{3})=0`
`⇔` \(\left[ \begin{array}{l}tan\ x+1=0\\tan\ x-\sqrt{3}=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}tan\ x=-1\\tan\ x=\sqrt{3}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{\pi}{4}+k\pi\ (k \in \mathbb{Z})\\x=\dfrac{\pi}{3}+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={-\frac{\pi}{4}+k\pi\ (k \in \mathbb{Z});\frac{\pi}{3}+k\pi\ (k \in \mathbb{Z})}`
b) `4sin^2 x-4cos\ x-1=0`
`⇔ 4(1-cos^2 x)-4cos\ x-1=0`
`⇔ 4-4cos^2 x-4cos\ x-1=0`
`⇔ 4cos^2 x+4cos\ x-3=0`
`⇔ (2cos\ x-1)(2cos\ x+3)=0`
`⇔` \(\left[ \begin{array}{l}2cos\ x-1=0\\2cos\ x+3=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}cos\ x=\dfrac{1}{2}\ (TM)\\cos\ x=-\dfrac{3}{2}\ (L)\end{array} \right.\)
`⇔ cos\ x=cos\ (\frac{\pi}{3})`
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={\frac{\pi}{3}+k2\pi\ (k \in \mathbb{Z});-\frac{\pi}{3}+k2\pi\ (k \in \mathbb{Z})}`
c) `\frac{1}{cos^2 x}-(3+\sqrt{3})tan\ x-1+\sqrt{3}=0`
`⇔ 1+tan^2 x-(3+\sqrt{3})tan\ x-1+\sqrt{3}=0`
`⇔ tan^2 x-(3+\sqrt{3})tan\ x+\sqrt{3}=0`
`⇔` \(\left[ \begin{array}{l}tan\ x=\dfrac{3+\sqrt{3}+\sqrt{12+2\sqrt{3}}}{2}\\tan\ x=\dfrac{3+\sqrt{3}-\sqrt{12+2\sqrt{3}}}{2}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l} x=arctan\ (\dfrac{3+\sqrt{3}+\sqrt{12+2\sqrt{3}}}{2})+k\pi\ (k \in \mathbb{Z})\\x=arctan\ (\dfrac{3+\sqrt{3}+\sqrt{12-2\sqrt{3}}}{2})+k\pi\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy `S={arctan\ (\frac{3+\sqrt{3}+\sqrt{12+2\sqrt{3}}}{2})+k\pi\ (k \in \mathbb{Z});arctan\ (\frac{3+\sqrt{3}+\sqrt{12-2\sqrt{3}}}{2})+k\pi\ (k \in \mathbb{Z})}`