`1+tanx=2\sqrt{2}sin(x+ π/4)` `(1)`
`ĐK: cosx\ne0⇔x\neπ/2 +kπ`
`(1)⇔ 1+sinx/cosx =2\sqrt{2}sin(x+ π/4)`
`⇔(cosx+sinx)/cosx=2\sqrt{2}sin(x+ π/4)`
`⇔cosx+sinx=2\sqrt{2}.cosx.sin(x+ π/4)`
`⇔\sqrt{2}sin(x+ π/4)-2\sqrt{2}.cosx.sin(x+ π/4)=0`
`⇔\sqrt{2}sin(x+ π/4)(1-2cosx)=0`
`⇔sin(x+ π/4)(1-2cosx)=0`
`⇔` $\left[\begin{matrix} sin(x+ \dfrac{π}{4})=0\\ (1-2cosx)=0\end{matrix}\right.$
`⇔` $\left[\begin{matrix}x= -\dfrac{π}{4}+kπ\\ x=\dfrac{π}{3}+k2π\\x= -\dfrac{π}{3}+k2π\end{matrix}\right.$
$\text{Vậy phương trình có ba họ nghiệm }$
$\left[\begin{matrix}x= -\dfrac{π}{4}+kπ\\ x=\dfrac{π}{3}+k2π\\x= -\dfrac{π}{3}+k2π\end{matrix}\right.$
$\text{Áp dụng}$
`cosx+sinx=\sqrt{2}sin(x+ π/4)`
