Answer :
As we need to find the most general solution for two different trigonometric equations.
Tan θ = -1 …(1)
and cos θ = 1/√2 …(2)
Most general value of θ is the common solution of both equation 1 and 2.
Let S1 represents the solution set of equation 1 and S2 be the solution set equation 2.
Let S represents the set of most general value of θ
∴ S = S1 ∩ S2
We know that, solution of tan x = tan α is given by –
x = nπ + α ∀ n ∈ Z
and solution of cos x = cos α is given by
x = 2mπ ± α ∀ m ∈ Z
From equation 1,we have –
tan θ = -1
⇒ tan θ = tan (-π/4)
∴ θ = nπ + (-π/4) = (nπ - π/4) ∀ n ∈ Z …(3)
From equation 2 we have -
cos θ = 1/√2
⇒ cos θ = cos π/4
⇒ θ = 2mπ ± π/4 ∀ m ∈ Z …(4)
From equation 3 we can infer that, solution lies either in 2nd quadrant (when n is odd) and 4th quadrant (when n is even)
From equation 3 we can infer that, solution lies either in 1st or 2nd quadrant irrespective of value of m.
∴ region of common solution is 4th quadrant and n is even in that case.
∴ common solution is given by-
θ = 2mπ - π/4 ∀ m ∈ Z.
The above solution is the most general solution for the given equations.
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