If 32tan8θ=2cos2α−3cosα  and 3cos2θ=1 , then the general value of α is

3cos2θ=1 ⇒tan2θ=1−cos2θ1+cos2θ=12 2cos2α−3cosα=32(116)⇒2cos2α−3cosα−2=0(2cosα+1)(cosα−2)=0⇒cosα=−12=cos2π3General solution is: α=2nπ±2π3;n∈z