Giaỉ các phương trình lượng giác sau:
1. sin(sinx)=0
2. sin(cosx)=0
3. \(\sqrt{3}\sin-\cos x=2cos3x\)
4. \(\sin2x=sin\left(2x-\dfrac{\pi}{2}\right)\)
5. \(4\cos\left(3\pi-2x\right)=\sqrt{2}\)
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NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 6 2021
1.
\(\Leftrightarrow2sinx.cosx+2cosx=0\)
\(\Leftrightarrow2cosx\left(sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=-1\end{matrix}\right.\)
\(\Leftrightarrow cosx=0\) (do \(cosx=0\Leftrightarrow sinx=\pm1\) bao hàm luôn cả pt \(sinx=-1\))
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
2.
\(\Leftrightarrow\left[{}\begin{matrix}2x-10^0=60^0+k360^0\\2x-10^0=120^0+n360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=35^0+k180^0\\x=65^0+n180^0\end{matrix}\right.\)
Do \(-120^0< x< 90^0\Rightarrow\left\{{}\begin{matrix}-120^0< 35^0+k180^0< 90^0\\-120^0< 65^0+n180^0< 90^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\n=\left\{-1;0\right\}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=35^0\\x=-115^0\\x=65^0\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 6 2021
3. Làm tương tự câu 2
4.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos\left(10x+\dfrac{4\pi}{5}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}cos\left(\dfrac{x}{2}-2\pi\right)\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}-2\pi\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)+cos\left(\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\dfrac{4\pi}{5}\right)=-cos\left(\dfrac{x}{2}\right)=cos\left(\pi-\dfrac{x}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\dfrac{4\pi}{5}=\pi-\dfrac{x}{2}+k2\pi\\10x+\dfrac{4\pi}{5}=\dfrac{x}{2}-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 6 2021
1.
Chắc đề là \(sin\left[\pi sin2x\right]=1?\)
\(\Leftrightarrow\pi.sin2x=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow sin2x=\dfrac{1}{2}+2k\) (1)
Do \(-1\le sin2x\le1\Rightarrow-1\le\dfrac{1}{2}+2k\le1\)
\(\Rightarrow-\dfrac{3}{4}\le k\le\dfrac{1}{4}\Rightarrow k=0\)
Thế vào (1)
\(\Rightarrow sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{6}+n2\pi\\2x=\dfrac{5\pi}{6}+m2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+n\pi\\x=\dfrac{5\pi}{12}+m\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 6 2021
2.
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+k2\pi\\\dfrac{\pi}{2}cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{\pi}{4}+k_12\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}+4k\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}+4k_1\end{matrix}\right.\) (2)
Do \(-1\le cos\left(x-\dfrac{\pi}{4}\right)\le1\Rightarrow\left\{{}\begin{matrix}-1\le\dfrac{1}{2}+4k\le1\\-1\le-\dfrac{1}{2}+4k_1\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}k=0\\k_1=0\end{matrix}\right.\)
Thế vào (2):
\(\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\\cos\left(x-\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\) chắc bạn tự giải tiếp được
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 2 2020
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
5 tháng 9 2021
1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)
⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)
⇔ 2cos2x - 5cosx + 2 = 0
⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên
2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)
⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0
⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)
⇒ sin4x + cos4x = 48.sin4x . cos4x
⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x
⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)
⇔ 1 - 2sin22x = 0
⇔ cos4x = 0
⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
5 tháng 9 2021
3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)
⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)
⇔ sin2x - sin22x - (1 + cos4x) = 0
⇔ sin2x - sin22x - 2cos22x = 0
⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0
⇔ sin22x + sin2x - 2 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)
⇔ sin2x = 1
⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
4, cos5x + cos2x + 2sin3x . sin2x = 0
⇔ cos5x + cos2x + cosx - cos5x = 0
⇔ cos2x + cosx = 0
⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
⇔ \(cos\dfrac{3x}{2}=0\)
⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)
⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)
Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)
⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}
Vậy các nghiệm thỏa mãn là các phần tử của tập hợp
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 10 2020
2.
\(\Leftrightarrow1-2cos^2\left(\frac{\pi}{4}-\frac{x}{2}\right)+sin\frac{x}{2}sinx-cos\frac{x}{2}sin^2x=0\)
\(\Leftrightarrow-cos\left(\frac{\pi}{2}-x\right)+sinx\frac{x}{2}sinx-cosx\frac{x}{2}sin^2x=0\)
\(\Leftrightarrow-sinx+sin\frac{x}{2}sinx-cos\frac{x}{2}sin^2x=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1-cos\frac{x}{2}sinx\right)=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1-2cos^2\frac{x}{2}sin\frac{x}{2}\right)=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1-2sin\frac{x}{2}\left(1-sin^2\frac{x}{2}\right)\right)=0\)
\(\Leftrightarrow sinx\left(2sin^3\frac{x}{2}-sin\frac{x}{2}-1\right)=0\)
\(\Leftrightarrow sinx\left(sin\frac{x}{2}-1\right)\left(2sin^2\frac{x}{2}+2sin\frac{x}{2}+1\right)=0\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 10 2020
1.
\(\Leftrightarrow sin2x-4sin\left(x+\frac{\pi}{4}\right)=5\)
Do \(\left\{{}\begin{matrix}sin2x\le1\\-4sin\left(x+\frac{\pi}{4}\right)\le4\end{matrix}\right.\) với mọi x
\(\Rightarrow sin2x-4sin\left(x+\frac{\pi}{4}\right)\le5\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}sin2x=1\\sin\left(x+\frac{\pi}{4}\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=-\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=-\frac{3\pi}{4}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 5 2019
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
3.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ