Question

Giải các phương trình sau:

a) Cot²x - (1 + \(\sqrt{3}\) )Cotx + \(\sqrt{3}\) = 0

b) 2Sin²2x + Sin2x - 1 = 0

c) tan²(x+1) + tan(x+1) - 2 = 0

d) Sin²x + Cosx +1 =0

e) 3cos²x - 5Sinx - 1 = 0

f) 2Cos2x - Cosx + 7 = 0

g) Sin4x + Cos4x = 2

h) Cosx - \(\sqrt{3}\)Sinx = -1

Answer 1

a. ĐKXĐ: ...

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

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

b.

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{1}{2}\end{matrix}\right.\)

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

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

Answer 2

c. ĐKXĐ: ...

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

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

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

d.

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

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

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

Answer 3

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

Answer 4

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.

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

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

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

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