Question

(vii) cos (120 ° + A) COS (120° - A) + cos (120° + A) cos A+ cos A (cos 120º - A) + 3/4 = 0
​

Answer 1

ANSWER

We have,

cos

3

x+cos

3

(120−x)+cos

3

(120+x)

Now, Using

cos3x=4cos

3

x−3cosx

cos

3

x=

4

cos3x+3cosx

Such that,

4

cos3x+3cosx

+

4

cos(360−3x)+3cos(120−x)

+

4

cos(360+3x)+3cos(120+x)

=

4

1

[cos3x+3cosx+cos(360−3x)+3cos(120−x)+cos(360+3x)+3cos(120+x)]

=

4

3

[cos3x+cosx+cos(120−x)+cos(120+x)]

=

4

3

[2cos2xcosx+2cos120cosx]

=

4

3

[2cos2xcosx+2(−

2

1

)cosx]

=

4

3

[cos3+cosx−cosx]

=

4

3

cos3x

Minimum value of cos3x=−1

So, minimum value of cos

3

x+cos

3

(120−x)+cos

3

(120+x)=

4

3

×−1=

4

−3

Hence, this is the answer.

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