The general solution of sin 2 θ sec θ+√3 tan θ=0 is

Question

The general solution of sin 2 θ sec θ+√3 tan θ=0 is (A) θ= n π+(-1) n +1 (π/3), θ= n π, n ∈ Z (B) θ=n π, n ∈ Z (C) θ= n π+(-1) n +1 (π/3), n ∈ Z (D) θ=( n π/2), n ∈ Z

Tardigrade Admin · Accepted Answer

The given equation can be written as ( sin 2 θ/ cos θ)+√3 tan θ=0 ⇒ tan θ sin θ+√3 tan θ=0 tan θ( sin θ+√3)=0 as sin θ ≠-√3 Hence, tan θ=0 ⇒ θ= n π, n ∈ Z

Q. The general solution of $sin^{2} θ sec θ + 3 tan θ = 0$ is

$θ = nπ + (- 1)^{n + 1} \frac{π}{3}, θ = nπ, n \in Z$

$θ = nπ, n \in Z$

$θ = nπ + (- 1)^{n + 1} \frac{π}{3}, n \in Z$

$θ = \frac{nπ}{2}, n \in Z$

The given equation can be written as
$\frac{s i n ^{2} θ}{c o s θ} + 3 tan θ = 0 \Rightarrow tan θ sin θ + 3 tan θ = 0$
$tan θ (sin θ + 3) = 0$ as $sin θ \neq = - 3$
Hence, $tan θ = 0$
$\Rightarrow θ = nπ, n \in Z$

Q. The general solution of sin2θsecθ+3​tanθ=0 is

θ=nπ+(−1)n+13π​,θ=nπ,n∈Z

θ=nπ,n∈Z

θ=nπ+(−1)n+13π​,n∈Z

θ=2nπ​,n∈Z