find the soln..........

i m getting sin 4x (2cosx -1) = 0

what after that????

4 Answers

1
sumit_kumar ·

but its not satisfying any of the options

106
Asish Mahapatra ·

sin6x + sin2x = sin4x
=> 2sin4xcos2x - sin4x=0
=> sin4x(2cos2x-1) = 0

=>4x=n\pi

OR

2x=2n\pi \pm \pi /3

1
nihal raj ·

sin6x = 2 sin(4x-2x)/2 . cos (4x+2x)/2

sin2(3x) = 2 sinx.cos3x

2 sin3x.cos3x= 2 sinx.cos3x

sin 3x = sinx

3sinx - 4 sin3x = sinx

3- 4sin2x=1

sin2x=1/2

sin x = + - 1/√2

looking at the option b satisfies above conditions...at n = + - 3

hence i think it should be option b)

1
sahil jain ·

b is correct

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