Q. 16

# Find the most gen

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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