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Question

If f(x)=sinx+cosx, then the number of solution of f(x)=[f(π10)], for x(0,4π) is
(where [.] represent greatest integer function)

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Solution

We know that,
f(π10)=sinπ10+cosπ10f(π10)=2sin(π10+π4)
Now,
sinπ4<sin(π10+π4)<sinπ212<sin(π10+π4)<11<2sin(π10+π4)<2
So,

Now the general solution will be,
sinx+cosx=1sin(π4+x)=12π4+x=2nπ+π4x=2nπ
Or
π4+x=2nπ+3π4x=2nπ+π2
Hence, x=π2,2π,5π2

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