Find at least one root of $\sin(t^2-1)=\sin(t)$

Question

I have solved one exercise then I have to solve the trigonometric equation at last to finish the exercise. The equation $\sin(t^2-1)=\sin(t)$

So I start to solve the following. Please give me a recommend or something to recover my solution:

$\begin{cases} t^2-1=t+2k\pi\\ t^2-1=\pi-t+2k\pi \end{cases} , k\in\mathbb{Z}$

Then I started to solve the quadratic equation using Discriminant.

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Did I do this in correct way ? Or something give me a hint about this.

Answer 1

Note that :

$sin(x) =sin(y) $$\Rightarrow$ $x=y+2k\pi $ or $x=\pi-x+2k\pi ( k \in \mathbb{Z} $)

So

$sin(t^2 - 1)=sin(t)$$\Rightarrow $$ t^2 - 1=t+2k\pi $ or $t^2 - 1=\pi-t+2k\pi $

if :

$t^2 - 1=t+2k\pi $

$\Rightarrow $ $t^2 - t-(1+2k\pi)=0$

$\triangle =5+8k\pi≠0$

If $\triangle <0 $ this equation have not a solution

If $\triangle >0 $the equation have two solutions

$t_1=\frac{1-\sqrt{5+8k\pi}} {2}$ and $t_2=\frac{1+\sqrt{5+8k\pi}}{2} $

Suppose :

$t^2 - 1=\pi-t+2k\pi $

$\Rightarrow $$ t^2 +t-(1+\pi+2k\pi)=0 $

$\triangle =1+4(\pi+1+2k\pi)≠0$

If$\triangle <0$ this equation have not a solution

If $\triangle >0$ this equation have two solutions

$t_1=\frac{-1-\sqrt{5+4(2k\pi+\pi)}}{2}$ And $t_2=\frac{-1+\sqrt{5+4(2k\pi+\pi)}}{2}$

Answer 2

Yes, you did this the correct way.

Rearranging the first equation to $t^2-t-2k\pi-1=0\\$ and then by the quadratic formula we have $$t=\frac{1\pm\sqrt{1+4(2k\pi+1)}}{2}=\frac{1}{2}\pm \frac{1}{2}\sqrt{5+8k\pi},\space\space k\in\mathbb Z$$ and similarly for the second quadratic $t^2+t-1-\pi-2k\pi=0$ we have $$t=\frac{-1\pm\sqrt{1+4(2k\pi+1+\pi)}}{2}=-\frac{1}{2}\pm \frac{1}{2}\sqrt{5+8k\pi+4\pi},\space\space k\in\mathbb Z$$