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\(\sin3x+\cos3x=3\sin x-4\sin^3x+4\cos^3x-3\cos x\)
\(=3\left(\sin x-\cos x\right)-4\left(\sin x-\cos x\right)\left(\sin^2x+\sin x\cos x+\cos^2x\right)=\left(\cos x-\sin x\right)\left(4\sin x\cos x+1\right)=\left(\cos x-\sin x\right)\left(1+2\sin2x\right)\)
\(\Leftrightarrow\sqrt{3}\cos x=\sin x\Leftrightarrow\sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
Bạn tự giải nốt, nhớ đối chiếu đkxd nhó
7.
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(\frac{\pi}{4}-x\right).sin\left(\frac{\pi}{4}+x\right)\ne0\\cos\left(\frac{\pi}{4}-x\right)cos\left(\frac{\pi}{4}+x\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow cos2x\ne0\)
Phương trình tương đương:
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{2}-\frac{\pi}{4}-x\right)}=cos^44x\)
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{4}-x\right)}=cos^24x\)
\(\Leftrightarrow sin^42x+cos^42x=cos^44x\)
\(\Leftrightarrow\left(sin^22x+cos^22x\right)^2-2sin^22x.cos^22x=cos^44x\)
\(\Leftrightarrow1-\frac{1}{2}sin^24x=cos^44x\)
\(\Leftrightarrow2-\left(1-cos^24x\right)=2cos^44x\)
\(\Leftrightarrow2cos^44x-cos^24x-1=0\)
\(\Leftrightarrow\left(cos^24x-1\right)\left(2cos^24x+1\right)=0\)
\(\Leftrightarrow cos^24x-1=0\)
\(\Leftrightarrow sin^24x=0\Leftrightarrow sin4x=0\)
\(\Leftrightarrow2sin2x.cos2x=0\Leftrightarrow sin2x=0\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)
1.
\(cos2x+5=2\left(2-cosx\right)\left(sinx-cosx\right)\)
\(\Leftrightarrow2cos^2x+4=4sinx-4cosx-2sinx.cosx+2cos^2x\)
\(\Leftrightarrow2sinx.cosx-4\left(sinx-cosx\right)+4=0\)
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-4t+4=0\)
\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
ĐKXĐ:...
Biến đổi đoạn trong ngoặc trước cho đỡ rối:
\(cos4x+sin2x=cos\left(3x+x\right)+sin\left(3x-x\right)\)
\(=cos3x.cosx-sin3x.sinx+sin3x.cosx-cos3x.sinx\)
\(=cosx\left(cos3x+sin3x\right)-sinx\left(cos3x+sin3x\right)\)
\(=\left(cosx-sinx\right)\left(cos3x+sin3x\right)\)
Thay vào phương trình:
\(\left(cosx-sinx\right)^2=2\left(sinx+cosx\right)+3\)
\(\Leftrightarrow1-2sinx.cosx=2\left(sinx+cosx\right)+3\)
Đặt \(sinx+cosx=a\Rightarrow-2sinx.cosx=1-a^2\)
\(2-a^2=2a+3\Rightarrow a=-1\Rightarrow sinx+cosx=-1\Rightarrow...\)
c/
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=cos3x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos3x\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=3x+k2\pi\\x+\frac{\pi}{3}=-3x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{12}+\frac{k\pi}{2}\end{matrix}\right.\)
d/
\(\Leftrightarrow\frac{1}{2}sin3x-\frac{\sqrt{3}}{2}cos3x=sin2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{3}=2x+k2\pi\\3x-\frac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{15}+\frac{k2\pi}{5}\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=sin\left(x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{6}\right)\)
\(\Rightarrow x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=sin\frac{\pi}{12}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=sin\frac{\pi}{12}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\frac{\pi}{12}+k2\pi\\x+\frac{\pi}{6}=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}.sin\left(3x-\dfrac{\pi}{4}\right)-\sqrt{2}.sin\left(5x-\dfrac{\pi}{3}\right)=0\Leftrightarrow sin\left(3x-\dfrac{\pi}{4}\right)=sin\left(5x-\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{4}+k2\pi=5x-\dfrac{\pi}{3}\\\pi-3x+\dfrac{\pi}{4}+k2\pi=5x-\dfrac{\pi}{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{19\pi}{96}+\dfrac{k\pi}{4}\end{matrix}\right.\); k\(\in Z\)
1.
\(\Leftrightarrow cos3x=-\frac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=40^0+k120^0\\x=-40^0+k120^0\end{matrix}\right.\)
\(\Rightarrow x=\left\{40^0;160^0;80^0\right\}\)
2.
Bạn coi lại đề, số \(-\sqrt{3}\) bên vế trái ko hề hợp lý, toán cho cấp 1 như vầy còn được chứ cấp 3 chắc ko ai cho đề kiểu vậy đâu
3.
\(\Leftrightarrow\sqrt{3}sin3x-cos3x=-sin5x-\sqrt{3}cos5x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=-\left(\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\right)\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(-5x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=-5x-\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\frac{4\pi}{3}+5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{48}+\frac{k\pi}{4}\\x=-\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
\(PT\Leftrightarrow\left\{{}\begin{matrix}sin3x+cos3x>=0\\2\cdot\left(sin3x+cos3x\right)^2=1+2\cdot sin6x+2\cdot sin2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}sin3x+cos3x>=0\\2+2\cdot sin6x=1+2\cdot sin6x+2\cdot sin2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}sin3x+cos3x>=0\left(1\right)\\sin2x=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)
(2): sin2x=1/2
=>2x=pi/6+k2pi hoặc 2x=5/6pi+k2pi
=>x=pi/12+kpi hoặc x=5/12pi+kpi
Khi x=pi/12+kpi thì:
\(sin3x+cos3x=sin\left(\dfrac{pi}{4}+3\cdot kpi\right)+cos\left(\dfrac{pi}{4}+3\cdot kpi\right)\)
Để sin 3x+cos3x>=0 thì k=2n
Khi x=5/12pi+kpi thì \(sin3x+cos3x=sin\left(\dfrac{5}{4}pi+3\cdot kpi\right)+cos\left(\dfrac{5}{4}pi+3\cdot k\cdot pi\right)\)
Để sin 3x+cos3x>=0 thì \(k=2n+1\)
=>Phương trình ban đầu sẽ có các nghiệm là: \(x=\dfrac{pi}{12}+2npi;x=\dfrac{17}{12}pi+2npi\)