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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/
\(\Leftrightarrow tan\left(2x-15^0\right)=1\)
\(\Rightarrow2x-15^0=45^0+k180^0\)
\(\Rightarrow x=30^0+k90^0\)
7.
Đặt \(\left|sinx+cosx\right|=\left|\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\right|=t\Rightarrow0\le t\le\sqrt{2}\)
Ta có: \(t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\frac{t^2-1}{2}\) (1)
Pt trở thành:
\(\frac{t^2-1}{2}+t=1\)
\(\Leftrightarrow t^2+2t-3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
Thay vào (1) \(\Rightarrow2sinx.cosx=t^2-1=0\)
\(\Leftrightarrow sin2x=0\Rightarrow x=\frac{k\pi}{2}\)
\(\Rightarrow x=\left\{\frac{\pi}{2};\pi;\frac{3\pi}{2}\right\}\Rightarrow\sum x=3\pi\)
6.
\(\Leftrightarrow\left(1-sin2x\right)+sinx-cosx=0\)
\(\Leftrightarrow\left(sin^2x+cos^2x-2sinx.cosx\right)+sinx-cosx=0\)
\(\Leftrightarrow\left(sinx-cosx\right)^2+sinx-cosx=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx-cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx-cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\x-\frac{\pi}{4}=-\frac{\pi}{4}+k\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=k\pi\\x=\frac{3\pi}{2}+k\pi\end{matrix}\right.\)
Pt có 3 nghiệm trên đoạn đã cho: \(x=\left\{\frac{\pi}{4};0;\frac{\pi}{2}\right\}\)
c/
\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x=0\)
\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx-\sqrt{3}cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\sinx=\sqrt{3}cosx\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
d/
\(\Leftrightarrow tan\left(3x-50^0\right)=-cot\left(x-30^0\right)\)
\(\Leftrightarrow tan\left(3x-50^0\right)=tan\left(x+60^0\right)\)
\(\Rightarrow3x-50^0=x+60^0+k180^0\)
\(\Rightarrow x=55^0+k90^0\)
a/
\(\Leftrightarrow sinx=2cosx\)
Nhận thấy \(cosx=0\) không phải nghiệm, pt tương đương:
\(\frac{sinx}{cosx}=2\Leftrightarrow tanx=2\)
\(\Leftrightarrow tanx=tana\) (với \(a\in\left(0;\frac{\pi}{2}\right)\) sao cho \(tana=2\))
\(\Rightarrow x=a+k\pi\)
b/
\(tan2x=cotx=tan\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow2x=\frac{\pi}{2}-x+k\pi\)
\(\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{3}\)
Xét \(x = {30^0} + k{90^0}\): Theo đề bài nên \(x = {30^0}(k \in \mathbb{Z})\) Chọn C