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a/
\(\Leftrightarrow\left(sinx-1\right)\left(sinx-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=4\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{2}+k2\pi\)
b/
\(\Leftrightarrow\left(cos2x-1\right)\left(2cosx-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sin3x-\frac{3}{4}\right)^2+\frac{7}{16}=0\)
Vế trái luôn dương nên pt vô nghiệm
\(4\left(cosx+1\right)+4\left(1-cos^2x\right)-5-3m=0\)
\(\Leftrightarrow-4cos^2x+4cosx+3=3m\)
Đặt \(f\left(x\right)=-4cos^2x+4cosx+3\)
\(f\left(x\right)=-\left(2cosx-1\right)^2+4\le4\)
\(f\left(x\right)=-4cos^2x+4cosx+8-5=4\left(cosx+1\right)\left(2-cosx\right)-5\ge-5\)
\(\Rightarrow-5\le f\left(x\right)\le4\)
\(\Rightarrow-5\le3m\le4\Rightarrow-\frac{5}{3}\le m\le\frac{4}{3}\)
a, ta có 2x + π/3 = 3π/4 +k2π hoặc 2x + π/3 = -3π/4 + k2π
=> x= 5π/24 + kπ hoặc x= -13π/24 +kπ
b, đề sai phải ko
c, cos22x - sin22x - 2sinx -1=0
<=> -2sin22x -2sin2x =0
<=> sin2x=0 hoặc sin2x=-1
<=> x=kπ hoặc x= π/2 + kπ ; x=-π/4 +kπ hoặc x=5π/8 + kπ
d, cos5xcosπ/4 - sin5xsinπ/4 = -1/2
cos( 5x + π/4 ) = -1/2
<=> x=π/12 +k2π/5 hoặc x= -11π/60 + k2π/5
f,4x+π/3=3π/10 -x +k2π hoặc 4x+π/3 = x - 3π/10 +k2π
<=> x =-π/150 + k2π/5 hoặc x = π/90 +k2π/3
1/ \(pt\Leftrightarrow\left(3cos^2x-sin^2x\right)\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow\left(\dfrac{3}{2}\left(1+cos2x\right)-\dfrac{1}{2}\left(1-cos2x\right)\right)\left(\dfrac{1}{2}\left(1+cos2x\right)-\dfrac{1}{2}\left(1-cos2x\right)\right)=0\)
\(\Leftrightarrow\left(2cos2x+1\right)cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
2/ \(pt\Leftrightarrow\left(sinx-1\right)\left(sin^2x+sinx+6\right)=0\)
\(\Leftrightarrow sinx=1\)
3/ \(pt\Leftrightarrow\dfrac{1-cos2x}{2}-4sin2x+\dfrac{7}{2}\left(1+cos2x\right)=0\)
\(\Leftrightarrow3cos2x-4sin2x=-4\)
\(\Leftrightarrow5\left(\dfrac{3}{5}cos2x-\dfrac{4}{5}sin2x\right)=-4\)
\(\Leftrightarrow cos\left(2x+arccos\dfrac{3}{5}\right)=-\dfrac{4}{5}\)
4,5 giải tương tự câu 3
d/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(\Leftrightarrow2\sqrt{2}\left(tanx+1\right)=\frac{3}{cos^2x}+2\)
\(\Leftrightarrow2\sqrt{2}tanx+2\sqrt{2}=3\left(1+tan^2x\right)+2\)
\(\Leftrightarrow3tan^2x-2\sqrt{2}tanx+5-2\sqrt{2}=0\)
Pt vô nghiệm
c/
\(\Leftrightarrow1-sin^2x+\sqrt{3}sinx.cosx-1=0\)
\(\Leftrightarrow\sqrt{3}sinx.cosx-sin^2x=0\)
\(\Leftrightarrow sinx\left(\sqrt{3}cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\\sqrt{3}cosx=sinx\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
e.
\(3\left(1-sin^2x\right)-5sinx-1=0\)
\(\Leftrightarrow-3sin^2x-5sinx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{3}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)
f.
\(2\left(2cos^2x-1\right)-cosx+7=0\)
\(\Leftrightarrow4cos^2x-cosx+5=0\)
Phương trình vô nghiệm
g.
\(\Leftrightarrow\sqrt{2}sin\left(4x+\frac{\pi}{4}\right)=2\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=\sqrt{2}>1\)
Phương trình vô nghiệm
h.
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
a) ta có : \(2sin^2x+3cos2x=0\Leftrightarrow2sin^2x+3\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow3-4sin^2x=0\Leftrightarrow sin^2x=\dfrac{3}{4}\Leftrightarrow sinx=\pm\dfrac{\sqrt{3}}{2}\)
th1 : \(sinx=\dfrac{\sqrt{3}}{2}\Leftrightarrow sinx=sin\dfrac{\pi}{3}\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\pi-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)th2 : \(sinx=\dfrac{-\sqrt{3}}{2}\Leftrightarrow sinx=sin\left(\dfrac{-\pi}{3}\right)\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\pi+\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{3}+k2\pi\end{matrix}\right.\)
vậy phương trình có 4 hệ nghiệm : \(x=\dfrac{\pi}{3}+k2\pi;x=\dfrac{2\pi}{3}+k2\pi;x=\dfrac{-\pi}{3}+k2\pi;x=\dfrac{4\pi}{3}+k2\pi\)
câu b bn làm tương tự cho quen nha
1. 4sin2x + 8cos2x-9=0
⇔ 4(sin2x+cos2x) + 4cos2x = 9
⇔ cos2x= \(\frac{9}{4}\)
⇔ cosx= \(\left[{}\begin{matrix}cosx=\frac{3}{2}\left(KTM\right)\\cosx=\frac{-3}{2}\left(KTM\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
2.
1-5sinx + 2cos2x=0
⇔1- 5sinx + 2(1-sin2x)=0
⇔ 2sin2x + 5sinx -3 =0
⇔\(\left[{}\begin{matrix}sinx=0,5\\sinx=-3\left(ktm\right)\end{matrix}\right.\)
Có sinx=0,5
⇔x=\(\left[{}\begin{matrix}x=\frac{\pi}{6}+2k\pi\\\frac{5\pi}{6}+2k\pi\end{matrix}\right.\left(k\in z\right)\)
Bạn sửa lại giúp mình câu 2 chỗ x đó là dấu ngoặc nhọn nhé, không phải dấu ngoặc vuông. Mình bị nhầm.