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# I want to know how to approach this problem..

Kaustubh Nayyar
27 Points
6 years ago
Just change the expression in terms of sin and cos (i.e. sina+cosa  and (sina)(cosa) )
Then replace sina.cosa as t and sina + cosa as (1+2t)1/2
This gives value of t . we have to find sin 2a = 2.sina.cosa=2t.
Kaustubh Nayyar
27 Points
6 years ago
Change the expression in terms of sina + cosa  and  sina.cosa
replace sina.cosa as t and sina+cosa  as (1 + 2t)½
(sina + cosa) + 1/(sina.cosa) + (sina + cosa)/(sina.cosa) = 7
(sina+cosa+1)(1/sina.cosa  + 1) = 8
(t + 1)(1+ (1+2t)1/2) = 8
this gives  2t3 – 44t2 + 18t = 0
t=0 or t2 – 22t + 18 = 0
put t = sina.cosa and multiply by 4
we get sin22a – 44 sin2a + 36 = 0
hence x = sinacosa
Kaustubh Nayyar
27 Points
6 years ago
x= sin2a