 # If for 0o  , e[(sin2x + sin4x + sin6x + ……)loge2] satisfies the quadratic equation x2-9x+8=0.Find the value of   sinx-cosx                          sinx +cosx 12 years ago

Dear student,

e[(sin2x + sin4x + sin6x + ……)loge2]

On solving we get 2^tan2x

x2-9x+8=0

this gives x=8,1

2^tan2x=1 and 2^tan2x=8

On critically examining we get x=pie/3

sin60-cos60/(sin60+cos60)=root 3-1/root 3+1

All the best.

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Sagar Singh

B.Tech, IIT Delhi

exp {(sin2x + sin4x + sin6x + ...... ∞) loge2

.log 2
1 sin x
sin x
2 e
2
e − ⇒ tan 2 x
eloge 2
⇒ tan2 x 2 satisfy x2 – 9x + 8 = 0 ⇒ x = 1, 8
∴ tan2 x 2 = 1 and tan2 x 2 = 8
⇒ tan2x = 0 and tan2x = 3
⇒ x = nπ and tan2x =
2
3
tan ?
?
?
???
π
and x = nπ ±
3
π
Neglecting x = nπ as 0 < x <
2
π
⇒ x =
3
π
∈ ?
?
?
??
? π
2
0,

cos x sin x
cos x
+
=
2
3
2
1
2
1
+
=
1 3
1
+
×
3 1
3 1

=
2
3 −1

cos x sin x
cos x
+
=
2
3 −1