Questions
General solution of is
a
2nπ+π6,2nπ−π2,n∈z
b
2nπ+π3,2nπ−π3,n∈z
c
2nπ,2nπ−2π3
d
2nπ+π4,2nπ+π3
detailed solution
Correct option is A
3cosx−sinx=1Dividing by (3)2+(−1)2=4=2⇒32cosx−12sinx=12⇒cosπ6cosx−sinπ6sinx=12⇒cosx+π6=cosπ3⇒x+π6=2nπ±π3,n∈zx=2nπ±π3−π6=2nπ+π6or 2nπ−π2,n∈zTalk to our academic expert!
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Number of values of and satisfying the equation is
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