cos−1(x)+cos−1(2x)+cos−1(3x)=π
cos−1(x)+cos−1(2x)=π−cos−1(3x)
cos−1(x(2x)−√1−x2√1−4x2)=π−cos−1(3x)
Taking cosθ on both sides , we get
2x2−√1−x2√1−4x2=−3x
2x2+3x=√1−x2√1−4x2
4x4+9x2+12x3=(1−x2)(1−4x2)
4x4+12x3+9x2=1−5x2+4x4
12x3+14x2−1=0
ax3+bx2+cx−1=0
Hence
a=12,b=14,c=0
Hence b−(a+c)
=14−12
=2