If cos ⁡3θ=cos⁡ 3α, then the value of sin θ can be given by

cos ⁡3θ=cos⁡ 3αPutting n = 0, 1, we have    3α=2nπ±3α∴     3θ=3α or −3α or 2π+3α or 2π−3α    θ=α or −α or 2π3+α or 2π3−αHence, options (a), (c), (d) are correctIf n=−1, then 3θ=−2π±3α⇒ θ=−2π3±α ⇒sin⁡ θ=sin⁡ −2π3±α=−sin⁡ 2π3±α=−sin ⁡π−π3±α=−sin ⁡π3±αHence, (b) is not correct.