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How to Solve 1 + sin 2x = ( sin 3x – cos 3x )whole square

 How to Solve 1 + sin 2x = ( sin 3x – cos 3x )whole square

Grade:12th pass

3 Answers

Bijoy Deka
12 Points
5 years ago
1 + sin 2x = ( sin 3x – cos 3x )2
1 + sin 2x =  sin2 3x +  cos2 3x – 2sin3x. cos3x
1 + sin 2x =  1 – 2sin3x. cos3x        (Since sin2 A +  cos2 A = 1)
 sin 2x =   – sin 6x                            ( sin 2A = 2sinAcosA)
sin 6x =  – sin 2x 
 sin 6x =   sin (π + 2x)                     ( Since  sin (π + A) = – sin A)
6x = nπ  – (π + 2x)                 or 6x = 2nπ  + (π + 2x)    n ∈ Z
6x + 2x = (n-1)π                       or 6x -2x = (2n+1)π    n ∈ Z
8x = (n-1)π                               or 4x = (2n +1)π    n ∈ Z
x = (n-1)π / 8                              or x = (2n +1)π /4   n ∈ Z
Ajay
209 Points
5 years ago
Wrong question 
Suppose x = 30 Degrees
Then LHS = 1 + sin60 =  1 + sqrt(3) /2
RHS = (sin 90 – cos 90)2 = 1
 
Hence question is wrong
Bijoy Deka
12 Points
5 years ago
This is not an Identity. It is an equation. Equations are called identities, if they are satisfied by all values of the unknown angles for which the functions are defined. 
 

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