Given the equation `sin(\frac{y}{4})-8cos(\frac{y}{6})=-3` . We have to solve for y on `0<y<50\pi`

Let `y=12x` . Then we have,

`sin(\frac{12x}{4})-8cos(\frac{12x}{6})=-3`

i.e. `sin(3x)-8cos(2x)=-3`

Bringing -3 to the left hand side of the equation, we get

`sin(3x)-8cos(2x)+3=0` ----->(1)

Now we know the 2 identities:

`sin(3x)=3sinx-4sin^3x` and,

`cos(2x)=1-2sin^2x`

Substituting these two identities in equation (1), we get

`3sinx-4sin^3x -8(1-2sin^2x)+3=0`

i.e. `3sinx-4sin^3x -8+16sin^2x+3=0`

`3sinx-4sin^3x+16sin^2x-5=0`

By rearranging, we can write

`-4sin^3x+16sin^2x+3sinx-5=0`

or,

`4sin^3x-16sin^2x-3sinx+5=0`

Now let `t=sinx` . Then, we can write the equation as:

`4t^3-16t^2-3t+5=0`

Here, we have to find the solution to this third degree polynomial.

Let us try to factorize the given polynomial.

`4t^3-16t^2-3t+5=0`

It can be written as:

`4t^3-14t^2-2t^2-10t+7t+5=0`

By rearranging, we can write

`4t^3-14t^2-10t-2t^2+7t+5=0`

`2t(2t^2-7t-5)-(2t^2-7t-5)=0`

i.e. `(2t-1)(2t^2-7t-5)=0`

Implies,

`2t-1 =0 ` or `2t^2-7t-5=0`

When `2t-1 =0 `

`t=\frac{1}{2}`

That means `sinx = \frac{1}{2}`

implies `x=sin^{-1}(\frac{1}{2})`

Now we know that `y=12x` . So,

`y=12sin^{-1}(\frac{1}{2})`

Now let us solve for `2t^2-7t-5=0`

Using quadratic formula:

`t= \frac{7\pm \sqrt{49-4(2)(-5)}}{2 \times 2}`

= `\frac{7\pm \sqrt{49+40}}{4}`

= `\frac{7\pm \sqrt{89}}{4}`

So, `t = \frac{7 +\sqrt{89}}{4}` or `t=\frac{7-\sqrt{89}}{4}`

i.e `t=4.12` or `t=-0.61`

i.e. `sinx = 4.12` or `sinx=-0.61`

or, `x=sin^{-1}(4.12) \ or \ x=sin^{-1}(-0.61)`

But 4.12 is outside the domain of sine inverse function.

So we have,

`y=12sin^{-1}(-0.61)`

Hence the solution is:

`y=12 sin^{-1}(\frac{1}{2})` and `y=12sin^{-1}(-0.61)`.

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