c) \(2cos2x-3cosx+1=0\Rightarrow2(2cos^2x-1)-3cosx+1=0\)

   \(\Rightarrow4cos^2x-3cosx-1=0\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{1}{4}\end{matrix}\right.\)

Bạn tự tìm x nhé.

f) \(cos2x+3sinx-2=0\Rightarrow1-2sin^2x+3sinx-2=0\)

   \(-2sin^2x+3sinx-1=0\) \(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

Bạn tự tìm x nhé.

a) \(2sim\left(2x-\dfrac{\pi}{3}\right)-1=0\Rightarrow sin\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}=sin\dfrac{\pi}{6}\)

    \(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\pi-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\) Bạn tự tìm x nhé.

d) \(\sqrt{3}sinx-cosx=\sqrt{2}\) \(\Rightarrow2sin\left(x-\dfrac{\pi}{6}\right)=\sqrt{2}\)

     \(\Rightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}=sin\dfrac{\pi}{4}\)

     \(\Rightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{6}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) Bạn tự tìm x nhé.

TL:

Take [math]a*x^2 + b*x +c = 0[/math]
Then
=>[math]x^2 + \frac{b}{a} x + \frac{c}{a} = 0[/math]
=>[math]x^2 + \frac{2b}{2a} x + \frac{c}{a} = 0[/math]
=>[math]x^2 + \frac{2b}{2a} x + (\frac{b}{2a})^2 + \frac{c}{a} - ( (\frac{b}{2a})^2 = 0[/math] -(1)

We have it in the form of [math]x^2 + 2px + p^2 + q = 0[/math]
which is [math](x+p)^2 + q = 0[/math]