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20 tháng 7 2022
1: \(\Leftrightarrow4\cdot\dfrac{1+\cos2x}{2}-6\cdot\dfrac{1-\cos2x}{2}+5\sin2x-4=0\)
\(\Leftrightarrow2+2\cos2x-3+3\cos2x+5\sin2x-4=0\)
\(\Leftrightarrow5\sin2x+5\cos2x=5\)
\(\Leftrightarrow\cos2x+\sin2x=1\)
\(\Leftrightarrow\sqrt{2}\cdot\sin\left(2x+\dfrac{\Pi}{4}\right)=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{4}=\dfrac{\Pi}{4}+k2\Pi\\2x+\dfrac{\Pi}{4}=\dfrac{3\Pi}{4}+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
2: \(\Leftrightarrow\sqrt{3}\cdot\dfrac{1+\cos2x}{2}+\sin2x-\sqrt{3}\cdot\dfrac{1-\cos2x}{2}-1=0\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}\cos2x+\sin2x+\sqrt{3}\cdot\dfrac{\cos2x-1}{2}-1=0\)
\(\Leftrightarrow\sin2x+\dfrac{\sqrt{3}}{2}\cos2x+\dfrac{\sqrt{3}}{2}\cos2x-\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}-2}{2}=0\)
\(\Leftrightarrow\sin2x+\sqrt{3}\cos2x=\dfrac{\sqrt{3}-\sqrt{3}+2}{2}=1\)
\(\Leftrightarrow\sin\left(2x+\dfrac{\Pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{\Pi}{3}=\dfrac{\Pi}{6}+k2\Pi\\2x+\dfrac{\Pi}{3}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{12}\Pi+k\Pi\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
T
11 tháng 7 2018
1. \(4\cos^2x-6\sin^2x+5\sin2x-4=0\)
\(\Leftrightarrow4\cos^2x-6\sin^2x+10\sin x\cos x-4\left(\cos^2x+\sin^2x\right)=0\)
\(\Leftrightarrow10\sin x\cos x-10\sin^2x=0\)
\(\Leftrightarrow10\sin x\left(\cos x-\sin x\right)=0\)
2. \(\sqrt{3}\cos^2x+2\sin x\cos x-\sqrt{3}\sin^2x-1=0\)
\(\Leftrightarrow\left(\sqrt{3}\cos^2x+\sin x\cos x\right)+\left(\sin x\cos x-\sqrt{3}\sin^2x\right)-1=0\)
\(\Leftrightarrow2\cos x\left(\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x\right)+2\sin x\left(\dfrac{1}{2}\cos x-\dfrac{\sqrt{3}}{2}\sin x\right)-1=0\)
\(\Leftrightarrow2\cos x.\cos\left(\dfrac{\Pi}{6}-x\right)+2\sin x.\sin\left(\dfrac{\Pi}{6}-x\right)-1=0\)
\(\Leftrightarrow\cos\dfrac{\Pi}{6}+\cos\left(2x-\dfrac{\Pi}{6}\right)+\cos\left(2x-\dfrac{\Pi}{6}\right)-\cos\dfrac{\Pi}{6}-1=0\)
\(\Leftrightarrow\cos\left(2x-\dfrac{\Pi}{6}\right)=\dfrac{1}{2}\)
3. \(2\sin^22x-3\sin2x\cos2x+\cos^22x=2\)
\(\Leftrightarrow2\sin^22x-3\sin2x\cos2x+\cos^22x-2\left(\sin^22x+\cos^22x\right)=0\)
\(\Leftrightarrow3\sin2x\cos2x+\cos^22x=0\)
\(\Leftrightarrow\cos2x\left(3\sin2x+\cos2x\right)=0\)
-TH1: ...
- TH2: \(\cos2x=-3\sin2x\) mà \(\cos^22x+\sin^22x=1\) suy ra ...
4. \(4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}=3\)
\(\Leftrightarrow4\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x+3\sin^2\dfrac{x}{2}-3\left(\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow\cos^2\dfrac{x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\dfrac{1+\cos x}{2}+\dfrac{1}{2}\sin x=0\)
\(\Leftrightarrow\cos x+\sin x=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
d.
\(\Leftrightarrow\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=0\)
\(\Leftrightarrow sin^2x-cos^2x=0\)
\(\Leftrightarrow-cos2x=0\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)
e. Đề thiếu
f.
\(\Leftrightarrow sin2x=\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\left(cos^2\frac{x}{2}+sin^2\frac{x}{2}\right)\)
\(\Leftrightarrow sin2x=cos^2\frac{x}{2}-sin^2\frac{x}{2}\)
\(\Leftrightarrow sin2x=cosx\)
\(\Leftrightarrow sin2x=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
a.
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\sqrt{2}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
b.
\(\Leftrightarrow sin2x=1\)
\(\Leftrightarrow2x=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
c.
\(\Leftrightarrow2sin2x.cos2x=-1\)
\(\Leftrightarrow sin4x=-1\)
\(\Leftrightarrow4x=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 8 2020
Nhân 2 vế với \(sin4x\) sau đó tách:
\(\frac{sin4x}{cosx}+\frac{sin4x}{sin2x}=\frac{2sin2x.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}=\frac{4sinx.cosx.cos2x}{cosx}+\frac{2sin2x.cos2x}{sin2x}\)
Rồi rút gọn
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
a.
\(\Leftrightarrow\left(1-sin^2x\right)\left(1+sin^2x\right)-\frac{5}{3}cos^4x=0\)
\(\Leftrightarrow cos^2x\left(1+sin^2x\right)-\frac{5}{3}cos^4x=0\)
\(\Leftrightarrow cos^2x\left(3+3sin^2x-5cos^2x\right)=0\)
\(\Leftrightarrow cos^2x\left(3+\frac{3}{2}-\frac{3}{2}cos2x-\frac{5}{2}-\frac{5}{2}cos2x\right)=0\)
\(\Leftrightarrow cos^2x\left(2-4cos2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
4 tháng 10 2020
a)bung hằng đẳng thức số 3 ra còn 5/3cos^4(x) giữ lại
Sau đó (1-sin^2(x)) là cos^2x sau đó rút nhân tử chung là cos^2(x) ra ta được
cos^2(x)(1+sin^2(x)-5/3cos^2(x))=0
Cho từng vế = 0 rr giải
b)rút sin x ra nhưng giữ thg cos2x lại rr rút nhân tử chung là cos2x ta đc
cos2x(1-sinx)=0
Cho từng vế =0 rr giải
c)chém 4cos^2(x) ở hai vế hai bên thì chỉ còn
cos3x+6cosx=0 <=> 4cos^3(x)+3cosx=0
Bấm máy tìm cosx
(sin2x - 4cos2x)(sin2x - 2sinx.cosx) = 2cos4x
⇔ (5sin2x - 4)(sin2x - sin2x) = 2cos4x
⇔ \(\left(\dfrac{5-5cos2x}{2}-4\right)\left(\dfrac{1-cos2x}{2}-sin2x\right)\)= 2cos4x
⇔ \(\dfrac{5-5cos2x-8}{2}.\dfrac{1-cos2x-2sin2x}{2}\) = 2cos4x
⇔ (5cos2x + 3)(cos2x + 2sin2x - 1) = 8cos4x
⇔ 5cos22x + 5cos2x.sin2x + 3cos2x + 6sin2x - 3 = 8cos4x
⇔ 5.\(\dfrac{1+cos4x}{2}\) + \(\dfrac{5}{2}sin4x\) + 3cos2x + 6sin2x - 3 = 8cos4x
⇔ \(\dfrac{5}{2}cos4x+\dfrac{5}{2}sin4x+3cos2x+6sin2x-\dfrac{1}{2}\) = 8cos4x
⇔ 5cos4x + 5sin4x + 6cos2x + 12sin2x - 1 = 16cos4x
VP = 16cos4x = 16 . \(\dfrac{\left(1+cos2x\right)^2}{4}\) = 4. (1 + cos2x)2
VP = 4 . (1 + 2cos2x + cos22x)
VP = 4 + 8cos2x + 4 . \(\dfrac{1+cos4x}{2}\)
VP = 6 + 8cos2x+ 2cos4x
Vậy 3cos4x + 5sin4x - 2cos2x + 12sin2x - 7 = 0
@Nguyễn Việt Lâm giúp em với