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Question

If cospθ+cosqθ=0 then the different values θ are in A.P with common difference 2π(p+q).

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Solution

cospθ+cosqθ=02cos(p+q2)θcos(pq2)θ=0
cos(p+q2)θ=0 or cos(pq2)θ=0
(p+q2)θ=(2n+1)π2 or (pq2)θ=(2n+1)π2
θ=(2n+1)πp+q or θ=(2n+1)πpq
Hence θ=πp+q,3πp+q...πpq,3πpq...A.P

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