Q11 part IV experts give solution

Q11 part IV experts give solution 11. Find the magnitude of angle A, if d: (i) 2 sin A cos A cos A— 2 sin j (ii) tan A—2 cos A tan A + 2 cog Au, J (iii) 2 cos2 A— 3 cos A + 1 = 0 (iv) 2 tan 3A cos 3A -- tan 3A + 1 s 2 un3A ue 12. Solve for x : (i) 2 cos 3x— 1 = 0 (ii) cos —e 1 (iii) sin (x + 100) = (iv) cos 2 (v) 2 cos (3x= 150) = I(vi) tan2 (vii) 3 tan2 (2x — 200) = 1 (viii) cos — + 100 2 (ix) sin2 x + sin2 300 (x) cos2 300 + cos2 x

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 )



Regards

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