The most general solution of tan θ=-1 and cos θ=(1/√2) is :

Question

The most general solution of tan θ=-1 and cos θ=(1/√2) is : (A) n π+(7 π/4), n ∈ I (B) n π+(-1)n (7 π/4), n ∈ I (C) 2 n π+(7 π/4), n ∈ I (D) n π+(-1)n (π/4), n ∈ I

Tardigrade Admin · Accepted Answer

tan θ=-1 ⇒ θ=(3 π/4), (7 π/4) text in [0,2 π] cos θ=(1/√2) θ=(π/4), (7 π/4) text in [0,2 π] common value is x=(7 π/4) general solution is 2 n π+(7 π/4), n ∈ I.

Q. The most general solution of $tan θ = - 1$ and $cos θ = \frac{1}{2}$ is :

$nπ + \frac{7 π}{4}, n \in I$

$nπ + (- 1)^{n} \frac{7 π}{4}, n \in I$

$2 nπ + \frac{7 π}{4}, n \in I$

$nπ + (- 1)^{n} \frac{π}{4}, n \in I$

$tan θ = - 1 \Rightarrow θ = \frac{3 π}{4}, \frac{7 π}{4} in [0, 2 π]$
$cos θ = \frac{1}{2} θ = \frac{π}{4}, \frac{7 π}{4} in [0, 2 π]$
common value is $x = \frac{7 π}{4}$
general solution is $2 nπ + \frac{7 π}{4}, n \in I$ .

Q. The most general solution of tanθ=−1 and cosθ=2​1​ is :

nπ+47π​,n∈I

nπ+(−1)n47π​,n∈I

2nπ+47π​,n∈I

nπ+(−1)n4π​,n∈I