Toán Giải phương trình: 3cotx-2=0 3tanx+4=0 6sinx-3=0 4cotx+2=0 cotx-1=0 sin(x + pi/2) sin 3x=0 22/07/2021 By Camila Giải phương trình: 3cotx-2=0 3tanx+4=0 6sinx-3=0 4cotx+2=0 cotx-1=0 sin(x + pi/2) sin 3x=0
Đáp án: $\begin{array}{l}a)3\cot x – 2 = 0\\ \Rightarrow \cot x = \dfrac{2}{3}\\ \Rightarrow x = arc\cot \dfrac{2}{3} + k\pi \\b)3\tan x + 4 = 0\\ \Rightarrow \tan x = – \dfrac{4}{3}\\ \Rightarrow x = \arctan \left( { – \dfrac{4}{3}} \right) + k\pi \\c)6\sin x – 3 = 0\\ \Rightarrow \sin x = \dfrac{1}{2}\\ \Rightarrow \left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\\d)4\cot x + 2 = 0\\ \Rightarrow \cot x = – \dfrac{1}{2}\\ \Rightarrow x = ar\cot \left( { – \dfrac{1}{2}} \right) + k\pi \\e)\cot x – 1 = 0\\ \Rightarrow \cot x = 1\\ \Rightarrow x = \dfrac{\pi }{4} + k\pi \\f)sin3x = 0\\ \Rightarrow 3x = k\pi \\ \Rightarrow x = \dfrac{{k\pi }}{3}\end{array}$ Trả lời
`a) 3cot x – 2 = 0` `<=> cot x = 2/3` `<=> x = ar“cc“ot (2/3) + kπ, k ∈ ZZ` `b) 3tan x + 4 = 0` `<=> tan x = -4/3` `<=> x = arctan (-4/3) + kπ, k ∈ ZZ` `c) 6sin x – 3 = 0` `<=> sin x = 1/2` `<=>` \(\left[ \begin{array}{l}x = \dfrac{π}{6} + k2π\\x = \dfrac{5π}{6} + k2π\end{array} \right.\) `(k ∈ ZZ)` `d) 4cot x + 2 = 0` `<=> cot x = -1/2` `<=> x = ar“cc“ot (-1/2) + kπ, k ∈ ZZ` `e) cot x – 1 = 0` `<=> cot x = 1` `<=> x = π/4 + kπ, k ∈ ZZ` `f) sin (x + π/2) = 0` `<=> x + π/2 = kπ` `<=> x = -π/2 + kπ, k ∈ ZZ` `g) sin 3x = 0` `<=> 3x = kπ` `<=> x = k(π)/3, k ∈ ZZ` Trả lời