Question

Giải các phương trình sau:

a, 2sin²x-(2+\(\sqrt{3}\))sinx+\(\sqrt{3}\)=0

b, \(\sqrt{3}cot^2x-4cotx+\sqrt{3}=0\)

c, 6sin²2x+sin2x-1=0

d, 4cos²x-11cosx+\(\sqrt{3}\)=0

e, sin²x+cosx+1=0

f, 2cos2x-4cosx=1

g, \(\frac{1}{sin^22x}\)-cot2x+1=0

h, tanx-\(\frac{1}{tanx}+2=0\)

Answer 1

@Nguyễn Việt Lâm giúp em với ạ

Answer 2

Ủa ko nhìn thấy bài này :)))

a.

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\) (pt bậc 2 có \(a+b+c=0\))

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

b.

\(\Leftrightarrow\left[{}\begin{matrix}cotx=\sqrt{3}\\cotx=\frac{1}{\sqrt{3}}\end{matrix}\right.\)

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

Answer 3

c.

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=arcsin\left(\frac{1}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{1}{3}\right)+k2\pi\end{matrix}\right.\)

d. Đề xấu quá, chắc bạn ghi nhầm dữ liệu

e.

\(\Leftrightarrow1-cos^2x+cosx+1=0\)

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

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

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

Answer 4

f.

\(\Leftrightarrow2\left(2cos^2x-1\right)-4cosx=1\)

\(\Leftrightarrow4cos^2x-4cosx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{3}{2}\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)

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

g. ĐKXĐ; ...

\(\Leftrightarrow1+cot^22x-cot2x+1=0\)

\(\Leftrightarrow cot^22x-cot2x+2=0\)

Pt vô nghiệm

h. ĐKXĐ: ...

\(tan^2x+2tanx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1-\sqrt{2}\\tanx=-1+\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{8}+k\pi\\x=-\frac{3\pi}{8}+k\pi\end{matrix}\right.\)