giải phương trình sinx cos2x = sin2x cos3x

Đáp án:

$\left[\begin{matrix} x=kπ\\ x=π/8+kπ/4\end{matrix}\right.$

Giải thích các bước giải:

 `sinx.cos2x=sin2x.cos3x`

`⇔sinx.cosx=2sinx.cosx.cos3x`

`⇔sinx(cos2x-2cosx.cos3x)=0`

`⇔`$\left[\begin{matrix} sinx=0(1)\\ cos2x-2cosx.cos3x=0(2)\end{matrix}\right.$

`(1)⇔sinx=0⇔x=kπ`

`(2)⇔2cos^2x-1=2cosx.cos3x`

`⇔2cos^2x-1=2cosx(4cos^3x-3cosx)`

`⇔2cos^2x-1=8cosx^4-6cos^2x`

`⇔8cos^4x-8cos^2x+1=0`

Đặt `cos^2x=t`

PT`⇔8t^2-8t+1=0`

`⇔`$\left[\begin{matrix} t=\frac{2+\sqrt[]{2}}{4}\\ t=\frac{2-\sqrt[]{2}}{4}\end{matrix}\right.$

`+)cos^2x=`$\frac{2+\sqrt[]{2}}{4}$`⇔`$\frac{1+cos2x}{2}$=$\frac{2+\sqrt[]{2}}{4}$

`⇔2+2cos2x=2+`$\sqrt[]{2}$

`⇔cos2x=`$\frac{\sqrt[]{2}}{2}$

`⇔2x=±π/4+k2π`

`⇔x=±π/8+kπ`

`+)

`cos^2x=`$\frac{2-\sqrt[]{2}}{4}$`⇔`$\frac{1+cos2x}{2}$=$\frac{2-\sqrt[]{2}}{4}$

`⇔2+2cos2x=2-`$\sqrt[]{2}$

`⇔cos2x=`$-\frac{\sqrt[]{2}}{2}$

`⇔2x=±(3π)/4+k2π`

Do `-1<cosx<1`

`⇒`$\left[\begin{matrix} x=kπ\\ x=π/8+kπ/4\end{matrix}\right.$