Bài 1:
a) 4sin23x + 2(\(\sqrt{3}\)+1) cos 3x - \(\sqrt{3}\)= 4
b) cos2x + 9cosx + 5 = 0
c) 4cos5(2 - 6x) + 16cos2(1 - 3x) =13
d)\(\frac{1}{cos^2x}-\left(3+\sqrt{3}\right)tanx-3+\sqrt{3}=0\)
e) \(\frac{3}{cosx}+tan^2x=9\)
f) 9 - 13cosx + \(\frac{4}{1+tan^2x}=0\)
g) \(\frac{1}{sin^2x}=cotx+3\)
h) \(\frac{1}{cos^2x}+3cot^2x=5\)
i) cos2x - 3cosx = 4cos2\(\frac{x}{2}\)
k) 2cos2x + tanx=\(\frac{4}{5}\)
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
\(\Rightarrow\left[{}\begin{matrix}tanx=\frac{3+\sqrt{3}-\sqrt{20+2\sqrt{3}}}{2}=tana\\tanx=\frac{3+\sqrt{3}+\sqrt{20+2\sqrt{3}}}{2}=tanb\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\end{matrix}\right.\)
5. ĐKXĐ: ...
\(\Leftrightarrow\frac{3}{cosx}+tan^2x+1-10=0\)
\(\Leftrightarrow\frac{1}{cos^2x}+\frac{3}{cosx}-10=0\)
Đặt \(\frac{1}{cosx}=a\Rightarrow\left|a\right|\ge1\)
\(a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{cosx}=2\\\frac{1}{cosx}=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\cosx=-\frac{1}{5}=cos\alpha\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\pm\alpha+k2\pi\end{matrix}\right.\)
6. ĐKXĐ \(cosx\ne0\)
\(9-13cosx+\frac{4}{1+tan^2x}=0\)
\(\Leftrightarrow9-13cosx+4cos^2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\frac{9}{4}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
7. ĐKXĐ \(sinx\ne0\)
\(\frac{1}{sin^2x}=cotx+3\)
\(\Leftrightarrow\frac{1}{sin^2x}-1-cotx-2=0\)
\(\Leftrightarrow\frac{cos^2x}{sin^2x}-cotx-2=0\)
\(\Leftrightarrow cot^2x-cotx-2=0\Rightarrow\left[{}\begin{matrix}cotx=-1\\cotx=2=cota\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=a+k\pi\end{matrix}\right.\)
8.
ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\frac{1}{cos^2x}-1+3cot^2x-4=0\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}+3cot^2x-4=0\)
\(\Leftrightarrow tan^2x+3cot^2x-4=0\)
Đặt \(cot^2x=t\ge0\)
\(\Rightarrow\frac{1}{t}+3t-4=0\Leftrightarrow3t^2-4t+1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cotx=\pm1\\cotx=\pm\frac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{4}+k\pi\\x=\pm\frac{\pi}{3}+k\pi\end{matrix}\right.\)
9.
\(\Leftrightarrow cos2x-3cosx=2\left(cosx+1\right)\)
\(\Leftrightarrow2cos^2x-1-3cosx=2cosx+2\)
\(\Leftrightarrow2cos^2x-5cosx-3=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=3\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{2\pi}{3}+k2\pi\)
10. ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(2cos2x+tanx=\frac{4}{5}\)
\(\Leftrightarrow4cos^2x-2+tanx=\frac{4}{5}\)
\(\Leftrightarrow\frac{4}{1+tan^2x}+tanx-\frac{14}{5}=0\)
Đặt \(tanx=t\)
\(\Rightarrow\frac{20}{1+t^2}+5t-14=0\)
\(\Leftrightarrow5t^3-14t^2+5t+6=0\)
\(\Leftrightarrow\left(t-2\right)\left(5t^2-4t-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{2+\sqrt{19}}{5}\\t=\frac{2-\sqrt{19}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanx=2=tana\\tanx=\frac{2+\sqrt{19}}{5}=tanb\\tanx=\frac{2-\sqrt{19}}{5}=tanc\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=a+k\pi\\x=b+k\pi\\x=c+k\pi\end{matrix}\right.\)
