# Thread: finding all the solutions in trig equations

1. ## finding all the solutions in trig equations

Hi folks,

I am trying to find the general solution to the following trig equation:

$tan 7\theta - tan(3\theta + \frac{\pi}{4}) = 0$

$\dfrac{sin 7\theta}{cos 7\theta} - \dfrac{sin(3\theta + \frac{\pi}{4})}{cos (3\theta + \frac{\pi}{4})} = 0$

$\dfrac{ sin 7\theta. cos (3\theta + \frac{\pi}{4}) - cos 7\theta. sin(3\theta + \frac{\pi}{4}) }{cos 7\theta. cos (3\theta + \frac{\pi}{4})} = 0$

Now, the numerator simplifies quite nicely to $sin (4\theta - \frac{\pi}{4}) = 0$ and it's quite easy to get the general solution. My question to the forum is about the denominator.

In equations with $\dfrac{numerator}{denominator} = 0$ we usually throw the denominator away (because it is zero) and then work only with the numerator and we find solutions for it.

If the numerator were, for example, $sin 4\theta. cos \theta = 0$ we would say $sin 4\theta = 0$ OR $cos \theta = 0$ and provide two sets of solutions.

So why don't we consider $cos 7\theta. cos(3\theta + \frac{\pi}{4}) = 0$?

thanks

2. ## Re: finding all the solutions in trig equations

Originally Posted by s_ingram
... In equations with $\dfrac{numerator}{denominator} = 0$ we usually throw the denominator away (because it is zero) and then work only with the numerator and we find solutions for it.
...
Where did you get that idea? We do not "throw the denominator away because it is zero" ... the denominator may not equal zero because the rational trig expression would be undefined. In fact, one should check the solutions determined by setting the numerator equal to zero to ensure none of them make the denominator zero.

$\dfrac{u}{v} = 0 \implies u=0$ and $v \ne 0$.

for example ...

$\dfrac{(x-1)(x-2)}{x^2-1} = 0$ has a single solution at $x=2$. The rational expression is undefined at $x= \pm 1$ even though the numerator equals zero at $x=1$.

3. ## Re: finding all the solutions in trig equations

thanks skeeter,

so you are saying that I should check that $cos 7\theta .cos(3\theta + \frac{\pi}{4}) \ne 0$ by finding solutions that do equal zero and eliminating them if they occur in the solutions for the numerator?

If I have this right, $\theta$ must not be $\frac{1}{7}(n\pi + \frac{\pi}{2})$ or $\frac{1}{3}(n\pi + \frac{\pi}{4})$. I assume I don't have to show these values unless they already occur in my solutions for the numerator?

4. ## Re: finding all the solutions in trig equations

both $7\theta$ and $\left(3\theta + \dfrac{\pi}{4}\right)$ cannot equal any odd-integer multiple of $\dfrac{\pi}{2}$.

5. ## Re: finding all the solutions in trig equations

When you solve only considering the numerator, you find that $\theta = \pi\left( \dfrac{4n+1}{16}\right)$. Now, to ensure that you only have valid solutions, you check to make sure that there are no values of $\theta$ where $7\theta = k\pi + \dfrac{\pi}{2}$ and no values such that $3\theta + \dfrac{\pi}{4} = k\pi + \dfrac{\pi}{2}$. Plugging in for $\theta$, you get:

$\dfrac{7\pi}{16}(4n+1) = k\pi + \dfrac{\pi}{2}$

Cancel out $\pi$, multiply out by 16:

$28n+7 = 16k+8 \Longrightarrow 28n-16k = 1$

But, the left hand side is the different of two even numbers while the right hand side is odd, which is not possible, so there are no integer values for $n,k$ satisfying that condition.

Next, check for $3\theta + \dfrac{\pi}{4} = k\pi + \dfrac{\pi}{2}$. Again, you will wind up with a difference of two even numbers equals an odd number when you plug in for $\theta$. So, there are no solutions that cause the numerator to be zero that also cause the denominator to be zero.

6. ## Re: finding all the solutions in trig equations

I suspect that, in "throw the denominator away because it is 0", "it" did not refer to the denominator but to the entire fraction. To solve $\displaystyle \frac{a}{b}= 0$ we can "throw the denominator away" because $\displaystyle \frac{a}{b}= 0$ if and only if $\displaystyle a= 0$.

7. ## Re: finding all the solutions in trig equations

Fantastic!

I have been struggling with this ever since skeeter pointed me in the right direction. I couldn't see a systematic way of doing it. I was trying values of n, so I had the general solution $\dfrac{5\pi}{16}$ for n = 1 and could see that my denominator solutions should not be $\dfrac{3\pi}{14}$ or $\dfrac{5\pi}{12}$ which they were not, but then there were all the other values of n to consider. So you have been a great help!

Many thanks SlipEternal.

8. ## Re: finding all the solutions in trig equations

Thanks HallsofIvy,

Exactly right. And I never thought further than that. But I think skeeter's point is also valid, because even if a = 0, if b is also 0, then the fraction is undefined and this has implications for the general solution as this post shows. I have never seen this issue covered in a maths textbook (maybe I should read more!) perhaps the authors think it is obvious, but SlipEternal's method seems to provide a complete answer. Do you agree?