Question

[Linhhh]

1) cosx\(^2\)+sinx=0

2) 2cos\(^2\)x-cos2x+cosx=0

3) sin\(^2\)x-3cos2x-2=0

4) tanx+\(\dfrac{2}{cotx}\)=0

Answer 1

[Nguyễn Việt Lâm]

3.

\(\dfrac{1}{2}-\dfrac{1}{2}cos2x-3cos2x-2=0\)

\(\Leftrightarrow-7cos2x-3=0\)

\(\Rightarrow cos2x=-\dfrac{3}{7}\)

\(\Rightarrow2x=\pm arccos\left(-\dfrac{3}{7}\right)+k2\pi\)

\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(-\dfrac{3}{7}\right)+k\pi\)

4.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(tanx+2tanx=0\)

\(\Rightarrow3tanx=0\)

\(\Rightarrow tanx=0\)

\(\Rightarrow x=k\pi\) (loại do ĐKXĐ)

Vậy pt đã cho vô nghiệm

Answer 2

[Nguyễn Việt Lâm]

1.

\(\Leftrightarrow1-sin^2x+sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1+\sqrt{5}}{2}>1\left(loại\right)\\sinx=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+k2\pi\\x=\pi-arcsin\left(\dfrac{1-\sqrt{5}}{2}\right)+k2\pi\end{matrix}\right.\) (\(k\in Z\))

2.

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

\(\Leftrightarrow cosx+1=0\)

\(\Leftrightarrow cosx=-1\)

\(\Leftrightarrow x=\pi+k2\pi\) (\(k\in Z\))