Given,"t2−9t+8=0, is satisfied by Now, alocos2x+cos4x+cos6x+...∞)logc2=blogca ∴e(cos2x+cos4x+...∞)log2 =2(cos2x+cos4x+∞)logee=2cos2x+cos4x+......∞ Here, cos2x+cos4x+...∞ are in GP, where a=cos2x ∴r=cos2x<1 ∴S∞=
a
1−r
∴S∞=
cos2x
1−cos2x
=
cos2x
sin2x
=cot2x ∴cos2x+cos4x+...∞=cot2x Now, 2cos2x+cos4x+...∞=2cot2x Now, roots of equation t2−9t+8=0, are (t−1)(t−8)=0 t=1,8 ⇒2cot2x=1 or 8 ⇒cot2x=0 or cot2x=3⇒cotx=0 or cotx=√3 \But here, 0<x<
π
2
. ∴cotx=0 not possible. Hence, cotx=√3 is only possible value. Now,
2sinx
sinx+√3cosx
Dividing numerator and denominator by sinx, we get
