Let S=[-π, (π/2))- -(π/2),-(π/4),-(3 π/4), (π/4) . Then the number of elements in the set A = θ ∈ S: tan θ(1+√5 tan (2 θ))=√5- tan (2 θ)

Question

Let S=[-π, (π/2))- -(π/2),-(π/4),-(3 π/4), (π/4) . Then the number of elements in the set A = θ ∈ S: tan θ(1+√5 tan (2 θ))=√5- tan (2 θ)  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

tan θ+√5 tan 2 θ tan θ=√5- tan 2 θ tan 3 θ=√5 θ=( n π/3)+(α/3) ; tan α=√5 Five solution

Q. Let $S = [- π, \frac{π}{2}) - {- \frac{π}{2}, - \frac{π}{4}, - \frac{3 π}{4}, \frac{π}{4}}$ .
Then the number of elements in the set
$A = {θ \in S : tan θ (1 + 5 tan (2 θ)) = 5 - tan (2 θ)}$

$tan θ + 5 tan 2 θ tan θ = 5 - tan 2 θ$
$tan 3 θ = 5$
$θ = \frac{nπ}{3} + \frac{α}{3}; tan α = 5$
Five solution

Q. Let S=[−π,2π​)−{−2π​,−4π​,−43π​,4π​}. Then the number of elements in the set A={θ∈S:tanθ(1+5​tan(2θ))=5​−tan(2θ)}

tanθ+5​tan2θtanθ=5​−tan2θ tan3θ=5​ θ=3nπ​+3α​;tanα=5​ Five solution

Q. Let $S = [- π, \frac{π}{2}) - {- \frac{π}{2}, - \frac{π}{4}, - \frac{3 π}{4}, \frac{π}{4}}$ .
Then the number of elements in the set
$A = {θ \in S : tan θ (1 + 5 tan (2 θ)) = 5 - tan (2 θ)}$

$tan θ + 5 tan 2 θ tan θ = 5 - tan 2 θ$
$tan 3 θ = 5$
$θ = \frac{nπ}{3} + \frac{α}{3}; tan α = 5$
Five solution