Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{2.sinx}{cosx}+\frac{cosx}{sinx}-4sinx.cosx-\frac{1}{2sinx.cosx}=0\) (Điều kiện \(x\ne\frac{k\pi}{2}\))
\(\Leftrightarrow\frac{4sin^2x+2cos^2x-8sin^2x.cos^2x-1}{2sinx.cosx}=0\)
\(\Leftrightarrow2sin^2x-8sin^2x.cos^2x+2\left(sin^2x+cos^2x\right)-1=0\)
\(\Leftrightarrow2sin^2x-8sin^2x.\left(1-sin^2x\right)+2-1=0\)
\(\Leftrightarrow8sin^4x-6sin^2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{4}\\sin^2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1-cos2x}{2}=\frac{1}{4}\\\frac{1-cos2x}{2}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\frac{1}{2}=cos\frac{\pi}{3}\\cos2x=0=cos\frac{\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{4}\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
giao với điều kiện \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\left(k\in Z\right)\)
1/ \(sinx=-\frac{1}{2}=sin\left(-\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
b/ \(cos=-\frac{\sqrt{2}}{2}=cos\left(\frac{3\pi}{4}\right)\)
\(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)
c/ \(tanx=\sqrt{3}=tan\left(\frac{\pi}{3}\right)\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
d/ \(cotx=0\Rightarrow x=\frac{\pi}{2}+k\pi\)
2/
a/ \(sin^2x+sinx-2=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-2\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
b/ \(cot^2x-2cotx-3=0\)
\(\Leftrightarrow\left(cotx+1\right)\left(cotx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cotx=-1\\cotx=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=arccot3+k\pi\end{matrix}\right.\)
3/ \(\Leftrightarrow1-cos2x+1-cos4x+1-cos6x=3\)
\(\Leftrightarrow cos2x+cos6x+cos4x=0\)
\(\Leftrightarrow2coss4x.cos2x+cos4x=0\)
\(\Leftrightarrow cos4x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\frac{2\pi}{3}+k2\pi\\2x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)
b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx
⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x
⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x
⇔ 4sin2x + (sinx + cosx) . sin2x = 0
⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)
⇔ sin2x = 0
c, 2cos3x = sin3x
⇔ 2cos3x = 3sinx - 4sin3x
⇔ 4sin3x + 2cos3x - 3sinx(sin2x + cos2x) = 0
⇔ sin3x + 2cos3x - 3sinx.cos2x = 0
Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình
Xét cosx ≠ 0 chia cả 2 vế cho cos3x ta được :
tan3x + 2 - 3tanx = 0
⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)
d, cos2x - \(\sqrt{3}sin2x\) = 1 + sin2x
⇔ cos2x - sin2x - \(\sqrt{3}sin2x\) = 1
⇔ cos2x - \(\sqrt{3}sin2x\) = 1
⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)
⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)
e, cos3x + sin3x = 2cos5x + 2sin5x
⇔ cos3x (1 - 2cos2x) + sin3x (1 - 2sin2x) = 0
⇔ cos3x . (- cos2x) + sin3x . cos2x = 0
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)
a. cos2x = 1-sin2x
b. cos2x = 2cos2x - 1
c. 2cosx.cos2x = 1 + cos2x + cos3x
=> 2cosx.cos2x = 2cos2x + 4cos3x - 3cosx
=> cosx(2.(2cos2x - 1) - 2cosx - 4cos2x +3) = 0
=> cosx( -2cosx + 1) = 0
=> cosx=0 hoặc cosx = -1/2
Đáp án D.