Question

[Bình Trần Thị]

giải phương trình : a) cosx×cos3x+sin2x×sin6x+sin4x×sin6x=0

                               b) cosx+cos3x+2cos5x=0

Answer 1

[Lightning Farron]

b) cosx+cos3x+2cos5x=0

<=>cosx+cos5x+cos3x+cos5=0

<=>2cos3x*cos2x+2cos4x*cos=0

<=>2[4cos³x-3cosx]*cos2x+2[2cos²2x-1]*cosx=0

<=>cosx[4cos²x-3]*cos2x+[2cos²2x-1]*cosx=0

<=>cosx[4cos²x-3]*cos2x+(2cos²x-1)=0

<=>cosx[2cos²x-cos2x(4cos²x-3)-1]=0

<=>cosx[4cos²x-2-(8cos⁴-x-10cos²x+3)-1]=0

<=>cosx[-8cos⁴x+14cos²x-6]=0

\(\Leftrightarrow\left[\begin{array}{nghiempt}cosx=0\\cos^2x=1\\cos^2x=\frac{3}{4}\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}cosx=0\\cosx=\pm1\\cosx=\pm\frac{\sqrt{3}}{2}\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{2}+k\pi\\x=k2\pi\\x=\pi+k2\pi\\x=\pm\frac{\pi}{6}+k2\pi\end{array}\right.\)

Answer 2

[Lightning Farron]

phần a đề có vấn đề r