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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