If

`sin^3 2x cos 6x + cos^3 2x sin 6x = (3/4) sin 8x`

Deduce the values of `a` for which the equation `sin^3 2x cos 6x + cos^3 2x sin 6x = a` is solvable.

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`sin^3 2x cos 6x + cos^3 2x sin 6x = a`

`sin^3 2x cos 6x + cos^3 2x sin 6x = (3/4)sin8x`

`(3/4)sin8x = a`

`sin8x = (4a)/3`

We know that for an angle A;

`-1<=sinA<=+1`

So in our question we can say;

`-1<=sin8x<=+1`

`-1<=(4a)/3<=+1`

`-3/4<=a<=3/4`

** So the range of values for `a` is **`a in [-3/4,3/4]`

**Sources:**

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