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Answer to Question #31368 in Trigonometry for Aparna

Question #31368
Plz show how to solve-
4sin x*sin 2x*sin 4x=sin 3x
Let&#039;s use the formula
sinx siny =1/2(cos(x-y) - cos(x+y))
So sin x * sin2x = 1/2 (cosx - cos3x)

We have: 4*sin x * sin 2x * sin 4x = 4 * 1/2 (cos x - cos3x) * sin 4x =
2*cos x *sin 4x &ndash; 2*cos 3x * sin 4x.

Using the formula sinx cosy = 1/2(sin(x+y) + sin(x-y)),we have:
2*cos x * sin 4x= sin 5x + sin 3x
and
2*cos 3x * sin4x = sin 7x + sin x

So 2*cos x * sin 4x &ndash; 2*cos 3x * sin 4x = sin 5x + sin 3x- sin 7x &ndash; sin x, which equals to the right side of initial equation sin 3x:

sin 5x + sin 3x- sin 7x &ndash; sin x = sin 3x
sin 5x - sin 7x&ndash; sin x = 0,
sin 5x - sin 7x= sin x,
-2*sin x * cos6x = sin x,
sin x * (2*cos6x + 1) = 0,
where from either sin x = 0 or cos 6x = -1/2.

The solution to sin x = 0 is x = pi * k, where k is anyinteger The solution to cos 6x = -1/2 is 6x = +/-2*pi / 3 + 2*pi*k or x = +/-pi / 9 + pi*k/3 where k is any integer

Answer: x = pi * k and x = +/-pi / 9 + pi*k/3 where k isany integer

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