Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c/ ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow tan^3x+1+tan^2x+4\sqrt{3}\left(1+tanx\right)=8+7tanx\)
\(\Leftrightarrow tan^2x\left(1+tanx\right)+\left(4\sqrt{3}-7\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left(tan^2x-7+4\sqrt{3}\right)\left(1+tanx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tan^2x=7-4\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=2-\sqrt{3}\\tanx=-2+\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=tan\left(-\frac{\pi}{4}\right)\\tanx=tan\left(\frac{\pi}{12}\right)\\tanx=tan\left(-\frac{\pi}{12}\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{12}+k\pi\\x=-\frac{\pi}{12}+k\pi\end{matrix}\right.\)
Bạn tự tìm x thuộc khoảng đã cho
b/
ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow tan^22x+1+tan^22x=7\)
\(\Leftrightarrow tan^22x=3\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=tan60^0\\tan2x=tan\left(-60^0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=60^0+k180^0\\2x=-60^0+k180^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-30^0+k180^0\end{matrix}\right.\)
Bạn tự tìm nghiệm thuộc khoảng đã cho nhé
3.
ĐKXĐ: ...
\(\Leftrightarrow tan^22x+\left(\frac{1}{cos^22x}+1\right)=8\)
\(\Leftrightarrow tan^22x+tan^22x=8\)
\(\Leftrightarrow tan^22x=4\)
\(\Rightarrow\left[{}\begin{matrix}tan2x=2\\tan2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=arctan\left(2\right)+k180^0\\2x=-arctan\left(2\right)+k180^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}arctan\left(2\right)+k90^0\\x=-\frac{1}{2}arctan\left(2\right)+k90^0\end{matrix}\right.\)
Nghiệm trên nhận các giá trị \(k=\left\{0;1;2;3\right\}\) ; nghiệm dưới nhận các giá trị \(k=\left\{1;2;3;4\right\}\)
1. ĐKXĐ: ...
\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=\frac{1}{tan\left(2x-\frac{\pi}{4}\right)}\)
\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=cot\left(2x-\frac{\pi}{4}\right)\)
\(\Leftrightarrow tan\left(x+\frac{\pi}{3}\right)=tan\left(\frac{3\pi}{4}-2x\right)\)
\(\Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{4}-2x+k\pi\)
\(\Rightarrow x=\frac{5\pi}{36}+\frac{k\pi}{3}\)
2.
ĐKXĐ: ...
\(\Leftrightarrow tan\left(x+1\right)=\frac{1}{cot\left(2x+3\right)}\)
\(\Leftrightarrow tan\left(x+1\right)=tan\left(2x+3\right)\)
\(\Leftrightarrow2x+3=x+1+k\pi\)
\(\Rightarrow x=-2+k\pi\)
c/
\(\Leftrightarrow tan\left(60^0-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Rightarrow60^0-x=-30^0+k180^0\)
\(\Rightarrow x=90^0+k180^0\)
d/
\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=-tan\left(\frac{\pi}{5}\right)\)
\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=tan\left(-\frac{\pi}{5}\right)\)
\(\Rightarrow3x+\frac{2\pi}{5}=-\frac{\pi}{5}+k\pi\)
\(\Rightarrow x=-\frac{\pi}{5}+\frac{k\pi}{3}\)
a/
\(\Leftrightarrow tan2x=-tan40^0\)
\(\Leftrightarrow tan2x=tan\left(-40^0\right)\)
\(\Rightarrow2x=-40^0+k180^0\)
\(\Rightarrow x=-20^0+k90^0\)
b/
\(\Leftrightarrow tan\left(2x-15^0\right)=1\)
\(\Rightarrow2x-15^0=45^0+k180^0\)
\(\Rightarrow x=30^0+k90^0\)
ĐKXĐ: ...
\(\Leftrightarrow tan\left(4x-\frac{\pi}{4}\right)=-tan\left(x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow tan\left(4x-\frac{\pi}{4}\right)=tanx\)
\(\Leftrightarrow4x-\frac{\pi}{4}=x+k\pi\)
\(\Rightarrow x=\frac{\pi}{12}+\frac{k\pi}{3}\)
a. ĐKXĐ; ...
\(cotx-2cotx=1\)
\(\Leftrightarrow cotx=-1\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
b. bạn coi lại đề
Nhưng có thể đề đúng là: \(2cos^2x-3cosx+1=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
Câu 2 bạn coi lại đề
3.
\(1+2sinx.cosx-2cosx+\sqrt{2}sinx+2cosx\left(1-cosx\right)=0\)
\(\Leftrightarrow sin2x-\left(2cos^2x-1\right)+\sqrt{2}sinx=0\)
\(\Leftrightarrow sin2x-cos2x=-\sqrt{2}sinx\)
\(\Leftrightarrow\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)=\sqrt{2}sin\left(-x\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=sin\left(-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{4}=-x+k2\pi\\2x-\frac{\pi}{4}=\pi+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
Bạn coi lại đề, xuất hiện 2 số hạng \(cos4x\) ở vế trái nên chắc là bạn ghi nhầm
5.
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=2cos^2\left(\frac{\pi}{4}-x\right)-1\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow sinx.sin2x-cosx.sin^22x=sin2x\)
\(\Leftrightarrow sin2x\left(sinx-cosx.sin2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\Leftrightarrow x=...\\sinx-cosx.sin2x-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
Ta có: \(tan\alpha\in\left(0;1\right)\) với mọi \(\alpha \in \left( {0;\dfrac{\pi }{4}} \right) \), do đó:
\(S = \underbrace {1 - \tan \alpha + {{\tan }^2}\alpha - {{\tan }^3}\alpha + ...}_{CSN\_lvh:{u_1} = 1,q = - \tan \alpha } = \dfrac{1}{{1 + \tan \alpha }} = \dfrac{{\cos \alpha }}{{\sin \alpha + \cos \alpha }} = \dfrac{{\cos \alpha }}{{\sqrt 2 \sin \left( {\alpha + \dfrac{\pi }{4}} \right)}}\)