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Solving trig equations with involving aperiodic functions

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**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.

Re: Solving trig equations with involving aperiodic functions

Where did your "principal domain" come from? I think it would more likely be $\displaystyle -\pi\leq 2x^2+x-1\leq \pi$.

Then knowing that $\displaystyle tan (x) = tan (x + \pi)$ you should be able to find four solutions.

Re: Solving trig equations with involving aperiodic functions

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