Question #5bc91

1 Answer
Dec 1, 2017

x_1=pi/12, x_2=(3pi)/4 and x_3=(17pi)/12

Explanation:

2(cos3x)^2-3sqrt2cos3x+2=0

(sqrt2cos3x)^2-3sqrt2cos3x+2=0

After using y=sqrt2cos2x transformation, this equation became,

y^2-3y+2=0 or (y-1)*(y-2)=0

Hence y_1=1 and y_2=2

For y=2, sqrt2*cos3x=2 or cos3x=sqrt2. This is impossible.

For y=1, sqrt2*cos3x=1 or cos3x=sqrt2/2. Hence 3x=pi/4+2pi*k or x=pi/12+(2pi*k)/3

Thus solutions of it are x_1=pi/12 (for k=0), x_2=(3pi)/4 (for k=1) and x_3=(17pi)/12 (for k=2)