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\(2sin2x+sinx.cosx-cos^2x+1=0\)
\(\Leftrightarrow4sin2x+2sinx.cosx-2cos^2x+2=0\)
\(\Leftrightarrow4sin2x+sin2x-cos2x=-1\)
\(\Leftrightarrow5sin2x-cos2x=-1\)
\(\Leftrightarrow\sqrt{26}\left(\dfrac{5}{\sqrt{26}}sin2x-\dfrac{1}{\sqrt{26}}cos2x\right)=-1\)
\(\Leftrightarrow cos\left(2x+arccos\dfrac{1}{\sqrt{26}}\right)=\dfrac{1}{\sqrt{26}}\)
\(\Leftrightarrow2x+arccos\dfrac{1}{\sqrt{26}}=\pm arccos\dfrac{1}{\sqrt{26}}+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-arccos\dfrac{1}{\sqrt{26}}+k\pi\end{matrix}\right.\)
a/
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow3tan^2x+8tanx+8\sqrt{3}-9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\sqrt{3}\\tanx=\frac{3\sqrt{3}-8}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k\pi\\x=arctan\left(\frac{3\sqrt{3}-8}{3}\right)+k\pi\end{matrix}\right.\)
b/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(tan^2x+2tanx-2=\frac{1}{2}\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^2x+4tanx-5=0\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=arctan\left(-5\right)+k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sinx+1\right)\left(1-2sin^2x-1\right)=0\)
\(\Leftrightarrow sin^2x\left(sinx+1\right)=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
sin3x + 1=2sin22x
<=> sin3x + 1 = 2\(\dfrac{1-cos4x}{2}\)
<=> sin3x + 1 = 1 - cos4x
<=> sin3x = -cos4x
<=> sin3x + cos4x = 0
<=> \(\dfrac{\sqrt{2}}{2}\)sin3x + \(\dfrac{\sqrt{2}}{2}\)cos4x = 0 (chia 2 vế cho \(\sqrt{2}\)).
<=> cos\(\dfrac{\pi}{4}\)sin3x + sin\(\dfrac{\pi}{4}\)cos4x = 0
<=> sin (3x+\(\dfrac{\pi}{4}\)) = 0
<=> sin(3x+\(\dfrac{\pi}{4}\)) = sin0
<=> \(\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=0+k2\pi\\3x+\dfrac{\pi}{4}=\pi-0+k2\pi\end{matrix}\right.\)(k\(\in\)Z)
<=>\(\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(k\(\in\)Z)
b) \(2sin^2x-3sinxcosx+cos^2x=0\)
\(\Leftrightarrow2tan^2x-3tanx+1=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{\pi}{4}\\tanx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{2}\right)+k\pi\end{matrix}\right.\left(k\in Z\right)\)
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
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)
Em ms hok cái này nên ko chắc lắm ạ :))
a/ \(\Leftrightarrow2\sin^2x.\cos x+3\sin x-4\sin^3x-4\cos^3x=0\)
Xét \(\sin^3x=0\) ko phải là nghiệm của PT
Xét \(\sin^3x\ne0\)
\(\Leftrightarrow2.\cot x+\frac{3}{\sin^2x}-4-4.\cot^3x=0\)
\(\Leftrightarrow4\cot^3x-3\cot^2x-2\cot x+1=0\)
Sau đó chị giải nghiệm là xong, thú thật e kém về phần gpt b3 trở lên nên sợ sai lắm :))
câu b khá là dài vì phải phân tích cos^3 2x nên ngày mai e giải nốt ạ :))
ĐK: \(x\ne k\pi\)
\(\dfrac{1+sin2x+cos2x}{1+cot^2x}=sinx.\left(sin2x+2sin^2x\right)\)
\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{cos^2x+sin^2x}{sin^2x}}=sinx.\left(2sinx.cosx+2sin^2x\right)\)
\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{1}{sin^2x}}=2sin^2x.\left(cosx+sinx\right)\)
\(\Leftrightarrow1+sin2x+cos2x=2cosx+2sinx\)
\(\Leftrightarrow1+2sinx.cosx+2cos^2x-1=2cosx+2sinx\)
\(\Leftrightarrow\left(cosx-1\right).\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(cosx-1\right).sin\left(x+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x+\dfrac{\pi}{4}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
a: =>sin2x+2*(1-cos2x)/2=2
=>sin2x-cos2x=1
=>căn 2*sin(2x-pi/4)=1
=>2x-pi/4=pi/4+k2pi hoặc 2x-pi/4=3/4pi+k2pi
=>x=pi/4+kpi hoặc x=pi/2+kpi
b: =>2*(1+cos2x)/2+1/2*sin2x-1/2(1-cos2x)=0
=>1+cos2x+1/2*sin2x-1/2+1/2cos2x=0
=>1/2*sin2x+3/2*cos2x=-1/2
=>sin(2x+a)=-cos(a)=cos(pi-a)
=>sin(2x+a)=sin(-pi/2+a)
=>2x+a=-pi/2+a+k2pi hoặc 2x+a=3/2pi-a+k2pi
=>x=-pi/4+kpi hoặc x=3/4pi-a+kpi