Re: Inverse functions and substitution

Solve: $\displaystyle 5\cos(x)+12\sin(x)=3$

Let $\displaystyle t=\tan\left(\frac{x}{2}\right)$ therefore we can write the equation as:

$\displaystyle 5\cdot \left(\frac{1-t^2}{1+t^2}\right)+12\cdot \sin\left(\frac{2t}{1+t^2}\right)-3=0$

$\displaystyle \Leftrightarrow \frac{5(1-t^2)+24t-3(1+t^2)}{1+t^2}=0$

$\displaystyle \Leftrightarrow 5(1-t^2)+24t-3(1+t^2)=0$

$\displaystyle \Leftrightarrow 5-5t^2+24t-3-3t^2=0$

$\displaystyle \Leftrightarrow -8t^2+24t+2=0$

Solutions:

$\displaystyle t_1=\frac{3-\sqrt{10}}{2}$ and $\displaystyle t_2=\frac{3+\sqrt{10}}{2}$

If we do the back-substitution:

$\displaystyle x=2\arctan\left(\frac{3+\sqrt{10}}{2}\right)+2k\pi$

$\displaystyle x=2\arctan\left(\frac{3-\sqrt{10}}{2}\right)+2k\pi$

(and $\displaystyle k \in \mathbb{Z}$)

So if we define the solution set:

$\displaystyle V=\{2\arctan\left(\frac{3-\sqrt{10}}{2}\right)+2k\pi,2\arctan\left(\frac{3+\ sqrt{10}}{2}\right)+2k\pi\}$

So we've solved the equation now there's nothing more to do and you don't have to worry about in which quadrant the function appears because that's the reason we define an infinite set of solutions.

If you want to be sure about the solution then take a look at wolphram alpha to compare.

Do you become the same result with the method The Chaz has given you?

Re: Inverse functions and substitution

Yes, I get the same answers with either method.

The confusion was the visualisation of what is actually happening behind the equations. How would I solve given a boundary of pi and -pi or [0,2pi)?

I think I get confused because I look at the answer and see tan so I imagine the graph of tan and wonder where the solutions are on it. Why is it 2kpi multiples if tan has period of just pi?

btw thanks again for all the help, it is more useful than you know :D

Re: Inverse functions and substitution

Quote:

Originally Posted by

**terrorsquid** I think I get confused because I look at the answer and see tan so I imagine the graph of tan and wonder where the solutions are on it. Why is it 2kpi multiples if tan has period of just pi?

In general if you have an equation:

$\displaystyle \tan(x)=\tan(a)$

The solutions of this equation are:

$\displaystyle x=a+k\pi$

But in this case we had something like this:

$\displaystyle \tan\left(\frac{x}{2}\right)=\tan(a)$

The solutions of this equation are:

$\displaystyle \frac{x}{2}=a+k\pi$

Multiplying both sides by 2 gives:

$\displaystyle x=2a+2k\pi$

Is this clear? ...