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Questions  

If 3sinx+4cosax=7 has at least one solution then the possible values of a

a
4m4n+1,m,n∈Z
b
4n+1π2,n∈Z
c
m4n+1,m,n∈Z
d
4n+14m,m,n∈Z

detailed solution

Correct option is A

We  have  sinx=1,cosa x=1⇒a x=2mπ,x=(4n+1)π2,m,n∈Z⇒(4n+1)π2=2mπa⇒a=4m4n+1,m,n∈z

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