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=0⇒2cos(p+q2)θcos(p−q2)θ=0 ⇒cos(p+q2)θ=0 or cos(p−q2)θ=0 ⇒(p+q2)θ=(2n+1)π2 or (p−q2)θ=(2n+1)π2 ⇒θ=(2n+1)πp+q or θ=(2n+1)πp−q Hence θ=πp+q,3πp+q...πp−q,3πp−q...→A.P