Number of points of discontinuity of $f(x)=\text{sgn}(\sin^2 x-\sin x-1)$ in $x\in(0,4\pi)$

Question

Number of points of discontinuity of $f(x)=\text{sgn}(\sin^2 x-\sin x-1)$ in $x\in(0,4\pi)$

Attempt for a solution

$$\text{sgn}(\sin^2 x-\sin x-1)$$

$$=\left\{\begin{matrix} -1\;\;, \;\sin^2x-\sin x-1<0\\\\ +0\;\;,\; \sin^2 x-\sin x-1=0 \\\\ 1\;\;,\; \sin^2x-\sin x-1>0 \end{matrix}\right.$$

but $\sin^2x -\sin x-1=\bigg(\sin x-\frac{1}{2}\bigg)^2-\frac{5}{4}$

How i approches that problem from that point . Help me please

Answer 1

For all $x \in \mathbb R$ we have

$$\sin^2x -\sin x+1=\bigg(\sin x-\frac{1}{2}\bigg)^2+\frac{3}{4}>0.$$

Hence $f(x)=1$ for all $x \in \mathbb R$ and therefore $f$ is continuous on $ \mathbb R$.

Answer 2

Hint: Use the third binomial formula $a^2-b^2=(a-b)(a+b)$:

$$\sin^2x -\sin x-1=\bigg(\sin x-\frac{1}{2}\bigg)^2-\frac{5}{4}=(\sin x-1/2-\sqrt{5}/2)(\sin x-1/2+\sqrt{5}/2).$$

Now, we know a product of two numbers is positive if both have the same sign. And we know that the product is negative if both numbers have opposite signs. It is zero if at least one of the numbers is equal to zero.

Can you solve it from here?