Pt bậc 1 theo sin,cos

[cau]

\[1: 5cos2x+6sin2x+13=0\]

\[2....\]

 

[beyumi_cute96]

1. 5cos2x+6sin2x+13=0

\[5cos2x+6sin2x+13=0\]

\[\Leftrightarrow \frac{5}{\sqrt{61}cos2x}+\frac{6}{\sqrt{61}sin2x}=\frac{-13}{\sqrt{61}}\]

Đặt:

\[\frac{5}{\sqrt{61}}=sin\alpha /\frac{6}{\sqrt{61}}=cos\alpha \]

\[pt\Leftrightarrow sin\alpha cos2x+cos\alpha sin2x=\frac{-13}{\sqrt{61}}\]

\[\Leftrightarrow sin\left(\alpha +2x \right)=\frac{-13}{\sqrt{61}}\]

\[\Leftrightarrow x=arcsin\left( \frac{-13}{2\sqrt{61}}\right) +\frac{k\Pi -\alpha }{2}\]

Hoặc

\[\Leftrightarrow x=\Pi -arcsin\left( \frac{-13}{2\sqrt{61}}\right) +\frac{k\Pi -\alpha }{2}\]



5cos2x+6sin2x+13=0
<=>5/căn61cos2x+6/căn61sin2x=-13/căn61
Đặt 5/căn 61=sinalpha, 6/căn 61=cosalpha
<=>sinalpha.cos2x+cosalpha.sin2x=-13/căn61
<=>sin(alpha+2x)=-13/căn 61
<=>x=arcsin-13/2 căn 61+kpi-alpha/2
x=pi-arcsin-13/2 căn 61+kpi-alpha/2