a/

\(\Leftrightarrow4cos^2\left(6x-2\right)+8\left(1+cos\left(6x-2\right)\right)-13=0\)

Đặt \(cos\left(6x-2\right)=a\Rightarrow\left|a\right|\le1\)

Pt trở thành:

\(4a^2+8\left(1+a\right)-13=0\)

\(\Leftrightarrow4a^2+8a-5=0\Rightarrow\left[{}\begin{matrix}a=\frac{1}{2}\\a=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cos\left(6x-2\right)=\frac{1}{2}\)

\(\Rightarrow6x-2=\pm\frac{\pi}{3}+k2\pi\)

\(\Rightarrow x=\frac{1}{3}\pm\frac{\pi}{18}+\frac{k\pi}{3}\)

b/

\(\Leftrightarrow2cos^2\left(x+75^0\right)-1+3sin\left(15^0-x\right)-1=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(90^0-15^0+x\right)-2=0\)

\(\Leftrightarrow2cos^2\left(x+75^0\right)+3cos\left(x+75^0\right)-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cos\left(x+75^0\right)=\frac{1}{2}\\cos\left(x+75^0\right)=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+75^0=60^0+k360^0\\x+75^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+k360^0\\x=-135^0+k360^0\end{matrix}\right.\)

c/

\(\Leftrightarrow\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x\right)+\left(\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx\right)=1\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)=1\)

\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow cos2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow1-2sin^2\left(x-\frac{\pi}{6}\right)+sin\left(x-\frac{\pi}{6}\right)-1=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)\left(1-2sin\left(x-\frac{\pi}{6}\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{6}\right)=0\\sin\left(x-\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=k\pi\\x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)

1.

\(4\left(1-cos^23x\right)+2\left(\sqrt{3}+1\right)cos3x-\sqrt{3}-4=0\)

Đặt \(cos3x=a\Rightarrow\left|a\right|\le1\)

\(\Rightarrow4\left(1-a^2\right)+2\left(\sqrt{3}+1\right)a-\sqrt{3}-4=0\)

\(\Leftrightarrow-4a^2+2\left(\sqrt{3}+1\right)a-\sqrt{3}=0\)

\(\Delta'=\left(\sqrt{3}+1\right)^2-4\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}a=\frac{-\sqrt{3}-1+\sqrt{3}-1}{-4}=\frac{1}{2}\\a=\frac{-\sqrt{3}-1-\sqrt{3}+1}{-4}=\frac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}cos3x=\frac{1}{2}\\cos3x=\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\pm\frac{\pi}{3}+k2\pi\\3x=\pm\frac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{9}+\frac{k2\pi}{3}\\x=\pm\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

2.

\(\Leftrightarrow2cos^2x-1+9cosx+5=0\)

\(\Leftrightarrow2cos^2x+9cosx+4=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow cosx=cos\left(\frac{2\pi}{3}\right)\)

\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)

3.

\(\Leftrightarrow4cos^5\left(2-6x\right)+8\left[cos\left(2-6x\right)+1\right]-13=0\)

Đặt \(cos\left(2-6x\right)=a\Rightarrow\left|a\right|\le1\)

\(\Rightarrow4a^5+8a-5=0\)

Bạn coi lại đề bài, pt bậc 5 ko nhẩm được nghiệm thì làm sao mà giải?

4.

ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow1+tan^2x-\left(3+\sqrt{3}\right)tanx-3+\sqrt{3}=0\)

\(\Leftrightarrow tan^2x-\left(3+\sqrt{3}\right)tanx-2+\sqrt{3}=0\)

\(\Delta=\left(3+\sqrt{3}\right)^2-4\left(-2+\sqrt{3}\right)=20+2\sqrt{3}\)

Chắc bạn lại nhầm hệ số nào đó, kết quả xấu kinh dị

\(\Rightarrow\left[{}\begin{matrix}tanx=\frac{3+\sqrt{3}-\sqrt{20+2\sqrt{3}}}{2}=tana\\tanx=\frac{3+\sqrt{3}+\sqrt{20+2\sqrt{3}}}{2}=tanb\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\end{matrix}\right.\)

1a.

Đặt \(5x+6=u\)

\(cos2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)

\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

1b.

Đặt \(2x+1=u\)

\(cos2u+3sinu=2\)

\(\Leftrightarrow1-2sin^2u+3sinu=2\)

\(\Leftrightarrow2sin^2u-3sinu+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)

2a.

\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)

\(\Leftrightarrow cos^2x+2cosx+1=0\)

\(\Leftrightarrow\left(cosx+1\right)^2=0\)

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\)

a/

\(\Leftrightarrow4sin^3x+6\sqrt{2}sinx.cosx-8sinx=0\)

\(\Leftrightarrow2sinx\left(2sin^2x+3\sqrt{2}cosx-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\2sin^2x+3\sqrt{2}cosx-4=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\left(1-cos^2x\right)+3\sqrt{2}cosx-4=0\)

\(\Leftrightarrow-2cos^2x+3\sqrt{2}cosx-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\sqrt{2}>1\left(l\right)\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)

b/

\(\Leftrightarrow4cos^3x+8sinx.cosx-7cosx=0\)

\(\Leftrightarrow cosx\left(4cos^2x+8sinx-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\4cos^2x+8sinx-7=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow4\left(1-sin^2x\right)+8sinx-7=0\)

\(\Leftrightarrow-4sin^2x+8sinx-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=\frac{3}{2}\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

ĐKXĐ; ...

\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cosx}{sinx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{sin^2x+cos^2x}{sinx.cosx}-5+\frac{3}{sin^22x}=0\)

\(\Leftrightarrow\frac{3}{sin^22x}+\frac{2}{sin2x}-5=0\)

Đặt \(\frac{1}{sin2x}=t\Rightarrow3t^2+2t-5=0\)

\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{5}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\end{matrix}\right.\)

a/

\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)

\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)

\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)

\(\Leftrightarrow-sin4x=sin^24x\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=k\pi\\4x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)

b/

\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)

\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)

\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)

\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)

\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)

a)

B1

b)

@Nguyễn Việt Lâm giúp em với ạ

a/

Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)

Pt trở thành:

\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)

\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)

\(\Leftrightarrow cos^2a+4sina-4=0\)

\(\Leftrightarrow1-sin^2a+4sina-4=0\)

\(\Leftrightarrow-sin^2a+4sina-3=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=1\)

\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)

b/

Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)

Pt trở thành:

\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)

\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)

\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)

\(\Leftrightarrow10sin^2a+4sina-14=0\)

\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)

\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)

\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)

c/

\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)

\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)

\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)

\(\Rightarrow x=...\)