# Juan was asked by his friend Mary to solve for y on 0<y<50 pi sine(y/4) - (8) cosine (y/6) = - 3. How can one solve for y in terms of pi ?

The solution to the given equation is y=12sin^{-1}(\frac{1}{2})  and y=12sin^{-1}(-0.61).

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

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).

Last Updated by eNotes Editorial on February 25, 2021