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e/
\(2cos^2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow1+cos2x+2cos^22x+4cos^32x-3cos2x=5\)
\(\Leftrightarrow2cos^32x+cos^22x-cos2x-2=0\)
\(\Leftrightarrow\left(cos2x-1\right)\left(2cos^22x+3cos2x+2\right)=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow x=k\pi\)
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):
\(\Leftrightarrow sinx-1-2sinx.cos^2x=0\)
\(\Leftrightarrow sinx-1-2sinx\left(1-sin^2x\right)=0\)
\(\Leftrightarrow2sin^3x-sinx-1=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2sin^2x+2sinx+1\right)=0\)
\(\Leftrightarrow...\)
1.
\(\Leftrightarrow\left(1-cos6x\right)cos2x+1-cos2x=0\)
\(\Leftrightarrow cos2x-cos2x.cos6x+1-cos2x=0\)
\(\Leftrightarrow\frac{1}{2}\left(cos8x-cos4x\right)-1=0\)
\(\Leftrightarrow2cos^24x-cos4x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-1\\cos4x=\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow4x=\pi+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
3.
Đặt \(\frac{x}{6}=t\Rightarrow\frac{1}{4}+cos^22t=\frac{1}{2}sin^23t\)
\(\Leftrightarrow1+4cos^22t=1-cos6t\)
\(\Leftrightarrow cos6t+4cos^22t=0\)
\(\Leftrightarrow4cos^32t+4cos^22t-3cos2t=0\)
\(\Leftrightarrow cos2t\left(4cos^22t+4cos2t-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2t=0\\cos2t=\frac{1}{2}\\cos2t=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{\pi}{4}+\frac{k\pi}{2}\\t=\pm\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}=\frac{\pi}{4}+\frac{k\pi}{2}\\\frac{x}{3}=\frac{\pi}{6}+k\pi\\\frac{x}{3}=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
d/
ĐKXĐ: ...
Biến đôi biểu thức vế trái trước:
\(1+tanx.tan\frac{x}{2}=1+\frac{sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}}=\frac{sinx.sin\frac{x}{2}+cosx.cos\frac{x}{2}}{cosx.cos\frac{x}{2}}=\frac{cos\left(x-\frac{x}{2}\right)}{cosx.cos\frac{x}{2}}=\frac{1}{cosx}\)
Do đó pt tương đương:
\(\sqrt{3}\left(1+tan^2x\right)-tanx-2\sqrt{3}=sinx.\frac{1}{cosx}\)
\(\Leftrightarrow\sqrt{3}tan^2x-2tanx-\sqrt{3}=0\)
\(\Rightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-\frac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Sử dụng kết quả biến đổi trên làm câu c sẽ lẹ hơn cách cũ
c/
ĐKXĐ: ...
\(\Leftrightarrow2cos^2x\left(1+tanx.tan\frac{x}{2}\right)=2cos^2x-4\)
\(\Leftrightarrow2cos^2x+2cos^2x.tanx.tan\frac{x}{2}=2cos^2x-4\)
\(\Leftrightarrow cos^2x.tanx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.tan\frac{x}{2}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=-2\)
\(\Leftrightarrow sinx.cosx.\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}.cos\frac{x}{2}}=-1\)
\(\Leftrightarrow cosx\left(\frac{1-cosx}{2}\right)=-1\)
\(\Leftrightarrow cos^2x-cosx-2=0\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pi+k2\pi\)
1.
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)
2.
\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)
\(\Leftrightarrow2cos^22x-cos2x=cos2x\)
\(\Leftrightarrow cos^22x-cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
3.
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow...\)
4.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
\(1+\sin\dfrac{x}{2}\sin x-\cos\dfrac{x}{2}\sin^2x=2\cos^2\left(\dfrac{\Pi}{4}-\dfrac{x}{2}\right)\)
\(\Leftrightarrow1+\sin\dfrac{x}{2}\sin x-\cos\dfrac{x}{2}\sin^2x=2\left(\dfrac{\sqrt{2}}{2}\cos\dfrac{x}{2}+\dfrac{\sqrt{2}}{2}\sin\dfrac{x}{2}\right)^2\)
\(\Leftrightarrow1+2\sin^2\dfrac{x}{2}\cos\dfrac{x}{2}-\cos\dfrac{x}{2}\left(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\right)^2=1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\)
\(\Leftrightarrow2\sin^2\dfrac{x}{2}\cos\dfrac{x}{2}-4\cos^3\dfrac{x}{2}\sin^2\dfrac{x}{2}-2\sin\dfrac{x}{2}\cos\dfrac{x}{2}=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\left(\sin\dfrac{x}{2}-2\sin\dfrac{x}{2}\cos^2\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\left(\sin\dfrac{x}{2}-2\sin\dfrac{x}{2}\left(1-\sin^2\dfrac{x}{2}\right)-1\right)=0\)
\(\Leftrightarrow2\sin\dfrac{x}{2}\cos\dfrac{x}{2}.\left(\sin\dfrac{x}{2}-1\right)\left(2\sin^2\dfrac{x}{2}+2\sin\dfrac{x}{2}+1\right)=0\)
3.
\(4sinx.cosx-2sinx+1-2cosx=0\)
\(\Leftrightarrow2sinx\left(2cosx-1\right)-\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
4.
\(cosx-sinx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\-4sinx.cosx=2t^2-2\end{matrix}\right.\)
Pt trở thành: \(t+2t^2-2-1=0\Leftrightarrow2t^2+t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{3}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}cos\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)
5.
\(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=sinx\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=x+k2\pi\\2x+\frac{\pi}{6}=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
6.
\(9sin^2x-5\left(1-sin^2x\right)-5sinx+4=0\)
\(\Leftrightarrow14sin^2x-5sinx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=arcsin\left(-\frac{1}{7}\right)+k2\pi\\x=\pi-arcsin\left(-\frac{1}{7}\right)+k2\pi\end{matrix}\right.\)
Câu 1:
\(\Leftrightarrow sinx.cos\frac{\pi}{3}-cosx.sin\frac{\pi}{3}+2\left(cosx.cos\frac{\pi}{6}+sinx.sin\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow sinx+\frac{1}{\sqrt{3}}cosx=0\)
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cosx\)
\(tanx+\frac{1}{\sqrt{3}}=0\Rightarrow tanx=-\frac{1}{\sqrt{3}}\Rightarrow x=\frac{\pi}{6}+k\pi\)
Câu 2:
\(\Leftrightarrow1-cos6x=1+cos2x\)
\(\Leftrightarrow-cos6x=cos2x\)
\(\Leftrightarrow cos\left(\pi-6x\right)=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi-6x+k2\pi\\2x=6x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
Câu 3:
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}-4\pi\right)+cos2x=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)+cos2x=1\)
\(\Leftrightarrow cos2x+cos2x=1\)
\(\Leftrightarrow cos2x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Câu 4:
\(\sqrt{2}\left(cosx.cos\frac{3\pi}{4}+sinx.sin\frac{3\pi}{4}\right)=1+sinx\)
\(\Leftrightarrow-cosx+sinx=1+sinx\)
\(\Leftrightarrow cosx=-1\Rightarrow x=\pi+k\pi2\)
Câu 5:
Giống câu 3, chắc bạn ghi nhầm đề
\(\Leftrightarrow8sinx.cosx\left(cos^2x-sin^2x\right)-\left(1+cos4x\right)=1\)
\(\Leftrightarrow4sin2x.cos2x-cos4x=2\)
\(\Leftrightarrow2sin4x-cos4x=2\)
\(\Leftrightarrow\frac{2}{\sqrt{5}}sin4x-\frac{1}{\sqrt{5}}cos4x=\frac{2}{\sqrt{5}}\)
\(\Leftrightarrow sin\left(4x+\alpha\right)=\frac{2}{\sqrt{5}}\)
\(\Leftrightarrow...\)
Nghiệm xấu quá, bạn tự giải nốt
ớ hình như phải là sin( 4x - α ) chớ