How to Solve 1 + sin 2x = ( sin 3x – cos 3x )whole square
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4
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Step-by-step explanation:
1 + sin 2x = ( sin 3x – cos 3x )²
1 + sin 2x = sin²3x + cos²3x – 2sin3x. cos3x
1 + sin 2x = 1 – 2sin3x. cos3x (∵ sin²A + cos²A = 1)
sin 2x = – sin 6x ( ∵sin 2A = 2sinAcosA)
sin 6x = – sin 2x
sin 6x = sin (π + 2x) (∵ 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
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1
Answer
nπ/4,nπ+π/4.
Just see watch the solution in the attachment below.
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