The equation `sin(y/4)-8cos(y/6)= -3` has to be solved for y.

Substitute `y = 12x` :

=> `sin((12x)/4)-8cos((12x)/6)= -3`

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

Sin 3x and cos 2x can be written in terms of sin x as follows:

`3*sin x - 4*sin^3x - 8(1-2*sin^2x) = -3`

`3*sin x - 4*sin^3x - 8 +16*sin^2x + 3 = 0`

Substitute z = sin x:

=> `3*z - 4*z^3 - 5 + 16z^2 = 0`

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

=> `2z(2z^2-7z-5)-1(2z^2-7z-5)=0`

=> `(2z-1)(2z^2-7z-5)=0`

=> `z = 1/2` and `z = 7/4+-sqrt89/4`

Now, z = sin x, but sin x lies in (-1, 1). It is defined for `z = 1/2` and `z = 7/4-sqrt89/4`

`sin x = 1/2` and `sinx = 7/4-sqrt89/4`

=> `x = sin^-1(1/2)` and `x = sin^-1(7/4-sqrt 89/4)`

As `y = 12x` ,

`y = 12*sin^-1(1/2)` and `y = 12sin^-1(7/4-sqrt 89/4)`

**The solution of the required equation is `y = 12*sin^-1(1/2) ` and `y = 12sin^-1(7/4-sqrt 89/4)`.**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now