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32tan power8 (a) = 2cos power2 (b) - 3cos (b) and 3cos2a =1 then find the general value of b.

32tan power8 (a) = 2cos power2 (b) - 3cos (b) and 3cos2a =1 then find the general value of b.

Grade:11

1 Answers

Aman
15 Points
4 years ago
32 tan^8⁡a=2cos^2⁡b-3 cos⁡b3 cos⁡2a=1cos⁡2a=1/3(1-tan^2⁡a)/(1+tan^2⁡a )=1/3tan^2⁡a=1/2tan^8⁡a=1/1632(1/16)=2cos^2⁡b-3 cos⁡b2=2cos^2⁡b-3 cos⁡b2 cos^2⁡b-3 cos⁡b-2=02 cos^2⁡b-4 cos⁡b+cos⁡b-2=02 cos⁡b (cos⁡b-2)+1( cos⁡b-2)=0(2 cos⁡b+1)(cos⁡b-2)=0either cos⁡b=-1/2cos⁡b=cos⁡〖2π/3〗b=2nπ±2π/3=2(nπ±π/3),n∈Zor cos⁡b=2But this is not possible as cos⁡b∈[-1,1]

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