Question

2tan3acos3a-tan3a+1=2cos3a; find the magnitude of a

Answer 1

Dear Student,

Please find below the solution to the asked query:

we have : 2 tan 3A cos 3A - tan 3A + 1 = 2 Cos 3A , So

2 tan 3A cos 3A - tan 3A - 2 Cos 3A + 1 = 0

tan 3A ( 2 Cos 3A - 1 ) - 1 ( 2 Cos 3A - 1 ) = 0

( tan 3A - 1 ) ( 2 Cos 3A - 1 ) = 0

So,

tan 3A - 1 = 0

tan 3A = 1

tan 3A = tan 45° ( As we know tan 45° = 1 )

And

tan 3A = tan π4 ( As we know 45° = π4 )

So,

3A = π4

A = π12

And

2 Cos 3A - 1 = 0

2 Cos 3A = 1

Cos 3A = 12

Cos 3A = Cos 60° ( As we know Cos 60° = 12 )
And

Cos 3A = Cos π3 ( As we know 60° = π3 )

3A = π3

A = π9
So,
We have

A = π9 and π12 ( Ans )