Question

tan pi by 7 +2 tan 2 pi by 7+4 tan 4 pi by 7 + 8 cot 8 pi by 7​

Answer 1

cosxcos2xcos3x=

4

1

(2cosxcos3x)cos2x=

2

1

(cos2x+cos4x)cos2x=

2

1

Letcos2x=2[cos4x=2cos

2

2x−1]

Now,

(t+2t

2

−1)t=

2

1

2t(2t

2

+t−1)=1

4t

3

+2t

2

−2t−1=0

2t

2

(2t+1)−(2t−1)=0

(2t

2

−1)(2t+1)=0

So,t=

2

−1

t=±

2

1

From

′

t

′

=

2

−1

cos2x=−

2

1

⇒2x=

3

2π

,

3

4π

Andfromt=±

2

1

cos2x=

2

1

⇒2x=

4

π

,

4

7π

x=

8

π

,

8

7π

Now,cos2x=

2

−1

2x=

4

3π

,

4

5π

⇒

8

π

,

3

π

,

8

3π

,

8

5π

,

3

2π

,

8

7π

thereare6solutions.

Hence,optionBiscorrectanswer.

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