1) Giai phuong trinh:
a) 3cos2x - 2sinx + 2 = 0
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NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2021
\(\Leftrightarrow2cos^22x+3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)=2\)
\(\Leftrightarrow4cos^22x-3cos2x-1=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=1\\cos2x=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=k2\pi\\2x=\pm arccos\left(-\dfrac{1}{4}\right)+k2\pi\\\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}arccos\left(-\dfrac{1}{4}\right)+k\pi\end{matrix}\right.\)
HP
20 tháng 9 2021
\(2cos^22x+3sin^2x=2\)
\(\Leftrightarrow-2\left(1-cos^22x\right)+3sin^2x=0\)
\(\Leftrightarrow-2sin^2x+3sin^2x=0\)
\(\Leftrightarrow sin^2x=0\)
\(\Leftrightarrow x=k\pi\)
HP
20 tháng 9 2021
Nguyễn Thái Sơn
\(\Leftrightarrow-2sin^22x+3sin^2x=0\)
\(\Leftrightarrow-2sin^22x+3sin^2x=0\)
\(\Leftrightarrow4sin^2x.cos^2x-3sin^2x=0\)
\(\Leftrightarrow sin^2x.\left(4cos^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=0\\cos^2x=\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=\pm\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)
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NT
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 9 2021
\(\Leftrightarrow\left[{}\begin{matrix}4x+10^0=x-20^0+k360^0\\4x+10^0=200^0-x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-30^0+k360^0\\5x=190^0+k360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-10^0+k120^0\\x=38^0+k72^0\end{matrix}\right.\) (\(k\in Z\))
28 tháng 5 2021
a/ \(\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-cos2x}{2}\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\dfrac{1-2cos^2x+1}{2}=\dfrac{2-2cos^2x}{2}=1-cos^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=\left(1-cosx\right)\left(1+cosx\right)\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1-cosx\right)\left(1+cosx\right)=0\)\(\Leftrightarrow\left(1+cosx\right)\left(2sinx-cosx-1+cosx\right)=0\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1+cosx=0\\2sinx-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=180^o\\x=30^o\end{matrix}\right.\)
28 tháng 5 2021
a) Đáp án: \(\left[{}\begin{matrix}cosx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))
Vậy...
b) \(3sin^2x+7cos2x-3=0\)
\(\Leftrightarrow3sin^2x+7\left(1-2sin^2x\right)-3=0\)
\(\Leftrightarrow11.sin^2x=4\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{2\sqrt{11}}{11}\\sinx=\dfrac{-2\sqrt{11}}{11}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\\x=\pi-arc.sin\dfrac{2\sqrt{11}}{11}+k2\pi\\x=arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\\x=\pi-arc.sin\dfrac{-2\sqrt{11}}{11}+k2\pi\end{matrix}\right.\) (\(k\in Z\)) (Dị quá,câu này e ko biết đ/a đúng hay sai đâu)
Vậy...
c)\(\dfrac{4.sin^2x+6.sin^2x-9-3.cos2x}{cosx}=0\) (đk: \(x\ne\dfrac{\pi}{2}+k\pi\),\(k\in Z\))
\(\Rightarrow10sin^2x-9-3\left(1-2.sin^2x\right)=0\)
\(\Leftrightarrow sin^2x=\dfrac{3}{4}\)\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{\sqrt{3}}{2}\\sinx=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)(\(k\in Z\)) (Thỏa mãn đk)
Vậy...
PT
1
CM
16 tháng 8 2017
3 cos 2 x - 2 sin x + 2 = 0 ⇔ 3 ( 1 - sin 2 x ) - 2 sin x + 2 = 0 ⇔ 3 sin 2 x + 2 sin x - 5 = 0 ⇔ ( sin x - 1 ) ( 3 sin x + 5 ) = 0 ⇔ sin x = 1 ⇔ x = π / 2 + k 2 π , k ∈ Z
\(3cos^2x-2sinx+2=0\)
\(\Leftrightarrow-3\left(1-cos^2x\right)-2sinx+5=0\)
\(\Leftrightarrow3sin^2x+2sinx-5=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(3sinx+5\right)=0\)
\(\Leftrightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
cos2x = 1 - sin2x chu