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sin2x can be rewritten

ya as 2sinxcosx but what next?

I don't have much idea next, leme see

yes thanks a lot thats absolutely fine! i got an idea to solve the question

Will you share that idea ?

2sinx=1; 2cosx=1 and further by the formulae sin theta =sin alpha and so on....

sinx and cosx both can not be 1/2 for the same value of x, you're logic is flawed.

Its quite complicated to solve this, are you sure you have the right ques ?

-sin2x !!

its easily solvable then

2cosx + 2sinx - 4sinx cosx = 1
2sinx(1 - 2cosx) = (1-2cosx)
=>(2sinx)(1- 2cosx) =0
BAZINGA !

wow!!! thanks....i was a bit confused....thanks a lot!

Glad to have helped.

Correction :

it'll be
(1-2sinx)(1- 2cosx) =0

ya i got that....k thanks bye gtg