Question

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Answer 1

Consider, y=exp[(sin

2

x+sin

4

x+sin

6

x+....+∞)log

e

2]

⇒y=exp[(

1−sin

2

x

sin

2

x

)log

e

2]=exp[tan

2

xlog

e

2]=2

tan

2

x

Since, y satisfies x

2

−9x+8=0, then

y=1,8

⇒2

tan

2

x

=2

0

,2

3

⇒tan

2

x=0,3

⇒tanx=0,±

3

Now, g(x)=

cosx+sinx

cosx

=

1+tanx

1

=1,

1±

3

1

=1,

2

−

3

−1

,

2

3

−1