# Solving trig equations with involving aperiodic functions

• July 13th 2014, 09:21 AM
maxpancho
Solving trig equations with involving aperiodic functions
Attachment 31263

C)

Principal domain:

$-\dfrac{\pi}{2}\leq 2x^2+x-1\leq \dfrac{\pi}{2}$

$-1.41099 \leq x \leq 0.91099$

Next:

$\sin(2x^2+x-1)=\dfrac{2}{5}$

$2x^2+x-1=\arcsin(\dfrac{2}{5})$

$x_1=-1.1265; x_2=0.62650$

Since $\sin(\theta)=-\sin(\theta)+\pi$ we get:

$\sin(-(2x^2+x-1)+\pi)=\dfrac{2}{5}$

$-2x^2-x+1+\pi=\arcsin(\dfrac{2}{5})$

Hence:

$x_3=1.1384; x_4=-1.6384$

Although, I wouldn't be able to find any more values of x by the methods I have been introduced to, since the functions is aperiodic.

By the same method I can find two values of x in D), but that's it. Although, I think they want me to do something else, rather than values for x.

P.S. Also, I haven't been introduced to solving trig equations with aperiodic functions, so it must be that they want to stretch me a little. So the tasks are probably rigged somehow, like in the case of C) I can find exactly four values with the methods I've been shown (well, as I understand), and that's what they want from me in the problem.

So, I guess it must be similar with the D) task, but I can't figure it out.
• July 18th 2014, 06:10 PM
Kiwi_Dave
Re: Solving trig equations with involving aperiodic functions
Where did your "principal domain" come from? I think it would more likely be $-\pi\leq 2x^2+x-1\leq \pi$.

Then knowing that $tan (x) = tan (x + \pi)$ you should be able to find four solutions.
• July 18th 2014, 07:22 PM
Karol95
Re: Solving trig equations with involving aperiodic functions