Animated Solution for Mathematics - Quadratic Equations: The equation $(\cos p-1)x^2+(\cos p)x+\sin p=0$ in the variable $x$, has real roots. Then $p$ can take any value in the interval

Imagine you are standing before a quadratic equation that looks like a standard $a{x}^2+bx+c=0$, but instead of simple integers, you see trigonometric functions dancing in the coefficients. The equation $(\cos p-1)x^2+(\cos p)x+\sin p=0$ is not just a math problem; it is a test of your ability to see through the complexity.

We are asked to find the interval for $p$ such that the roots are real. The gatekeeper of real roots is the discriminant, $D=b^2-4ac$. For any quadratic equation to have real roots, this discriminant must be non-negative, $D\ge 0$. This is our bedrock.

Let us extract our coefficients with precision. Comparing our equation to the standard form, we identify $a=\cos p-1$, $b=\cos p$, and $c=\sin p$.

Now, we substitute these into the discriminant formula:

This looks a bit messy, doesn't it? But let us simplify it with care. By distributing the negative sign into the bracket, we transform $(\cos p-1)$ into $-(1-\cos p)$.

This small, elegant algebraic shift turns our expression into:

Why did we do this? Because now, the structure of the expression is laid bare.

Look at the two components of our discriminant. The first term is ${\cos }^{2}p$. Since it is a square, it is always non-negative, ${\cos }^{2}p\ge 0$.

The second term involves $(1-\cos p)$. We know from the fundamental range of the cosine function that $-1\le \cos p\le 1$, which means $1-\cos p$ is always non-negative, $1-\cos p\ge 0$.

We have a sum of two non-negative terms, but there is a multiplier: $4\sin p$. For the entire discriminant $D$ to be guaranteed as non-negative, we need the term $4\sin p(1-\cos p)$ to be non-negative. Since $4$ is positive and $(1-\cos p)$ is non-negative, the burden of proof falls on $\sin p$. We need $\sin p\ge 0$.

Now, we turn to the unit circle. Where is the sine function positive? It is positive in the first and second quadrants, which corresponds to the interval $(0,\pi )$.

In this region, $\sin p>0$. When $\sin p>0$, our discriminant $D$ becomes a sum of non-negative terms, ensuring $D\ge 0$.

This confirms that for any $p$ in the interval $(0,\pi )$, the equation will indeed have real roots. We have navigated the algebra, respected the trigonometry, and arrived at the solution with clarity. The interval $(0,\pi )$ is our answer, a beautiful testament to the harmony between quadratic theory and the unit circle.