Câu hỏi:
Giúp em bài tập về nhà Toán lớp 11 câu hỏi như sau: tan2x=-tanx , cot3x=-cot2x , cos4x=-cos2x , zin3x=-sin2x
Trả lời 2:
Gia Sư Hoàng Khang gữi câu trả lời dành cho bạn:
a. $tan2x=-tanx$
$ĐK: \left\{ \begin{array}{l}cos2x \ne 0\\cosx \ne 0\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}x \ne \dfrac{\pi}{4}+k\dfrac{\pi}{2}\\x \ne \dfrac{\pi}{2}+k\pi \end{array} \right.$
$\Leftrightarrow 2x=-x+k\pi (k \in Z)$
$\Leftrightarrow x=k\dfrac{\pi}{3} (k \in Z)$
b. $cot3x=-cot2x$
$ĐK: \left\{ \begin{array}{l}sin3x \ne 0 \\sin2x \ne 0\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}x \ne k\dfrac{\pi}{3}\\x\ne k\dfrac{\pi}{2}\end{array} \right.$
$\Leftrightarrow 3x=-2x+k\pi (k \in Z)$
$\Leftrightarrow x=k\dfrac{\pi}{5} (k \in Z)$
c. $cos4x=-cos2x$
$\Leftrightarrow cos4x+cos2x=0$
$\Leftrightarrow 2cos3x.cosx=0$
$\Leftrightarrow \left[ \begin{array}{l}2cos3x=0\\cosx=0\end{array} \right. (k \in Z)$
$\Leftrightarrow \left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\dfrac{\pi}{3}\\x=\dfrac{\pi}{2}+k\pi\end{array} \right. (k \in Z)$
d. $sin3x=-sin2x$
$\Leftrightarrow \left[ \begin{array}{l}3x=-2x+k2\pi=0\\3x=\pi+2x+k2\pi\end{array} \right. (k \in Z)$
$\Leftrightarrow \left[ \begin{array}{l}x=k\dfrac{2\pi}{5} \\x=\pi+k2\pi\end{array} \right. (k \in Z)$
Trả lời 1:
Gia Sư Hoàng Khang gữi câu trả lời dành cho bạn: Tan2x=-tan x /
2x =(-x) +kπ/
x=kπ/3(k€z) .//////
Cot3x =-cot2x /
3x=(-2x) +kπ/
X=kπ/5. //////
Cos4x=-cos2x/
Cos4x=cos(π+2x) hoặc cos(π-2x) /
4x=π+2x+k2π
4x=-π-4x+k2π/
x=π/2+kπ
x=-π/8+kπ/2.///////
Sin3x=-sin 2x/
3x=(-2x) +k2π
3x=π+2x+k2π/
x=kπ/3
x=π+k2π/
