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Question
If cospθ+cosqθ=0, then the differentvalues of θ are in A.P. with a common difference
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Solution
The correct option is A 2π/(p±q)
Here, we can solve the equation as:
cospθ+cosqθ=0⇒cospθ=−cosqθ⇒cospθ=cos(π−qθ)Thus, the general solution of the above equation can be given as:pθ=2nπ±(π−qθ), n∈Z.⇒(p∓q)θ=(2n+1)π, n∈Z.⇒θ=(2n+1)π(p∓q), n∈Z.Now, putting values of n, we getvalues of θ=π(p∓q),3π(p∓q), . . clearly these solutions form an A.P. with common difference=2π(p±q).
Here, we can solve the equation as:
cospθ+cosqθ=0⇒cospθ=−cosqθ⇒cospθ=cos(π−qθ)Thus, the general solution of the above equation can be given as:pθ=2nπ±(π−qθ), n∈Z.⇒(p∓q)θ=(2n+1)π, n∈Z.⇒θ=(2n+1)π(p∓q), n∈Z.Now, putting values of n, we getvalues of θ=π(p∓q),3π(p∓q), . . clearly these solutions form an A.P. with common difference=2π(p±q).