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32tan power8 (a) = 2cos power2 (b) - 3cos (b) and 3cos2a =1 then find the general value of b.
32 tan^8a=2cos^2b-3 cosb3 cos2a=1cos2a=1/3(1-tan^2a)/(1+tan^2a )=1/3tan^2a=1/2tan^8a=1/1632(1/16)=2cos^2b-3 cosb2=2cos^2b-3 cosb2 cos^2b-3 cosb-2=02 cos^2b-4 cosb+cosb-2=02 cosb (cosb-2)+1( cosb-2)=0(2 cosb+1)(cosb-2)=0either cosb=-1/2cosb=cos〖2π/3〗b=2nπ±2π/3=2(nπ±π/3),n∈Zor cosb=2But this is not possible as cosb∈[-1,1]
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