If the solutions for $\theta$ from the equation $\sin^2 \theta-2\sin\theta+\lambda = 0$ lie in $\cup_{n\in\Bbb{z}}$...

Question

If the solutions for $\theta$ from the equation $\sin^2 \theta - 2\sin\theta + \lambda = 0$ lie in $\cup_{n \in \Bbb{z}}(2n\pi - \frac{\pi}{6}, (2n + 1)\pi + \frac{\pi}{6})$. Then find the possible set values of $\lambda$.

My problems with the question.

1. Firstly, i do not know how to start the question.
2. Secondly, I have always been having doubt with the exact meaning of $\cup_{n \in \Bbb{z}}$. So please help me clarify

Answer 1

Observe that

$\sin\left(2n\pi-\dfrac\pi6\right)=\sin\left(-\dfrac\pi6\right)=-\sin\dfrac\pi6=?$

$\sin\left((2n+1)\pi+\dfrac\pi6\right)=\sin\left(\pi+\dfrac\pi6\right)=-\sin\dfrac\pi6=?$

As $\sin x$ is increasing near $2n\pi-\dfrac\pi6$ and decreasing near $(2n+1)\pi+\dfrac\pi6,$

we need $\sin\theta>-\dfrac12\implies\sin\theta-1>-\dfrac32\implies(\sin\theta-1)^2<\left(-\dfrac32\right)^2$

Now $\lambda=1-(\sin\theta-1)^2-1>1-\left(-\dfrac32\right)^2$

Answer 2

You can ignore $\sin$ for a while by setting $x = \sin \theta$ so your equation becomes:

$x^2 - 2 x + \lambda = 0$

which should be easy to solve and, depending on $\lambda$ may have $0$, $1$, or $2$ solutions. You can now solve for $\theta$ by using $\sin^{-1}$ but remember that multiple solutions will be possible. You now need to check for which values of $\lambda$ you get the desired set.

I am not familiar with that syntax either. I expect that it just means the set containing each of those values (the expression) for all integer values of $n$. Something like this is to be expected since there are multiple possible solutions to $\sin^{-1} \theta$ for a specific $\theta$.

Answer 3

Every elementary equation for the sine has infinitely many solutions (or none): if $-1\le x\le 1$, then the equation $\sin\theta=x$ admits the solutions $$ \theta=\arcsin x+2n\pi, \qquad \theta=\pi-\arcsin x+2n\pi $$ The two families of solutions reduce to one when $x=1$ or $x=-1$.

Note that $-\pi/2\le\arcsin x\le\pi/2$ and $\pi/2\le\pi-\arcsin x\le3\pi/2$.

Your quadratic equation splits into two elementary equations (or one if the discriminant is $0$) and the condition is that the solutions fall in the stated union of intervals \begin{multline} \dots\cup (-4\pi-\tfrac{\pi}{6},-3\pi+\tfrac{\pi}{6})\cup (-2\pi-\tfrac{\pi}{6},-\pi+\tfrac{\pi}{6})\cup (-\tfrac{\pi}{6},\pi+\tfrac{\pi}{6})\\ \cup (2\pi-\tfrac{\pi}{6},3\pi+\tfrac{\pi}{6})\cup (4\pi-\tfrac{\pi}{6},\pi+\tfrac{\pi}{6})\cup \dotsb \end{multline}

The only interval among these that lies inside $(-\pi/2,3\pi/2)$ is the one for $n=0$, namely $(-\pi/6,\pi+\pi/6)$.

This just amounts to saying that, if $\sin\theta=x_1$ and $\sin\theta=x_2$ are the two elementary equations we get, we need \begin{gather} -\frac{\pi}{6}<\arcsin x_1<\frac{7\pi}{6} \qquad -\frac{\pi}{6}<\pi-\arcsin x_1<\frac{7\pi}{6} \\ -\frac{\pi}{6}<\arcsin x_2<\frac{7\pi}{6} \qquad -\frac{\pi}{6}<\pi-\arcsin x_2<\frac{7\pi}{6} \end{gather}

Since $a<\pi-b<c$ is equivalent to $\pi-c<b<\pi-a$, the first two conditions are actually the same and so for the other two conditions. Since, by definition, $\arcsin x\le\pi/2$, $$ -\frac{\pi}{6}<\arcsin x_1<\frac{7\pi}{6} $$ amounts to saying that $x_1>\sin(-\pi/6)=-1/2$. The same for $x_2$.

Thus the problem has been reduced to finding the values of $\lambda$ so that the roots of $x^2-2x+\lambda$ both satisfy $x>-1/2$.

Rewrite the equation as $x^2-2x+1+\lambda-1=0$, that is, $(x-1)^2=1-\lambda$. Since we want $x>-1/2$, we need $x-1>-3/2$, that is, $(x-1)^2<9/4$ which becomes $$ 1-\lambda<9/4 $$ or $\lambda>-5/4$.

Note that in the case the equation has no solution, or a solution is $>1$, there is no problem, because the (non existent) solution of $\sin\theta=x$ will satisfy any condition.