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 S_{1} represents the solution set of equation 1 and S_{2} be the solution set equation 2.

Let S represents the set of most general value of θ

∴ S = S_{1} ∩ S_{2}

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 2^{nd} quadrant (when n is odd) and 4^{th} quadrant (when n is even)

From equation 3 we can infer that, solution lies either in 1^{st} or 2^{nd} quadrant irrespective of value of m.

∴ region of common solution is 4^{th} 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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