Animated Solution for Mathematics - Trigonometry: The number of values of $x$ in the interval $\left(\frac{\pi }{4},\frac{7\pi }{4}\right)$ for which $14{\text{cosec}}^{2}x-2{\sin }^{2}x=21-4{\cos }^{2}x$ holds, is ______.

Welcome, fellow explorers of mathematics! Today, we are standing at the edge of a trigonometric forest. The equation before us, $14{\csc }^{2}x-2{\sin }^{2}x=21-4{\cos }^{2}x$, might look like a tangled mess of cosecants, sines, and cosines.

Our mission is to find the number of values of $x$ in the interval $\left(\frac{\pi }{4},\frac{7\pi }{4}\right)$ that satisfy this equality. Let us begin.

Simplifying the Chaos

The first rule of trigonometry is to find a common language. We know the fundamental identities: ${\cos }^{2}x=1-{\sin }^{2}x$ and ${\csc }^{2}x=\frac{1}{{\sin }^{2}x}$.

By substituting these into our equation, we can express everything in terms of ${\sin }^{2}x$:

Expanding the right-hand side, we get $21-4+4{\sin }^{2}x$, which simplifies to $17+4{\sin }^{2}x$. Now, our equation looks much cleaner:

By moving the $-2{\sin }^{2}x$ to the right, we consolidate our terms: $\frac{14}{{\sin }^{2}x}-6{\sin }^{2}x=17$. We have successfully tamed the jungle!

The Quadratic Transformation

Now, let's make this even easier. Let $p={\sin }^{2}x$. Our equation becomes a simple algebraic one: $\frac{14}{p}-6p=17$.

To clear the fraction, we multiply the entire equation by $p$, leading us to $14-6p^2=17p$. Rearranging this into the standard quadratic form, we get:

We are looking for two numbers that multiply to $6\times (-14)=-84$ and add up to $17$. Those numbers are $21$ and $-4$.

Splitting the middle term: $6p^2+21p-4p-14=0$. Grouping the terms, we get $3p(2p+7)-2(2p+7)=0$, which factors into $(3p-2)(2p+7)=0$.

This gives us two potential roots: $p=\frac{2}{3}$ or $p=-\frac{7}{2}$.

The Geometric Reality

We defined $p={\sin }^{2}x$. Since the square of any real number is non-negative, and the sine function is bounded between $-1$ and $1$, ${\sin }^{2}x$ must lie in the interval $[0,1]$.

Thus, we must reject $p=-\frac{7}{2}$. We are left with ${\sin }^{2}x=\frac{2}{3}$, which means $\sin x=\pm \sqrt{\frac{2}{3}}$.

Now, let's visualize this on the sine curve within our interval $\left(\frac{\pi }{4},\frac{7\pi }{4}\right)$. The value $\sqrt{\frac{2}{3}}\approx 0.816$.

The left boundary of our interval is $\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\approx 0.707$. Since $0.816>0.707$, the horizontal line $y=0.816$ intersects the sine curve twice within the region where $\sin x>0.707$.

Similarly, $-\sqrt{\frac{2}{3}}\approx -0.816$. The right boundary is $\sin \left(\frac{7\pi }{4}\right)=-\frac{1}{\sqrt{2}}\approx -0.707$. Since $-0.816<-0.707$, the line $y=-0.816$ also intersects the sine curve twice within the region where $\sin x<-0.707$.

By carefully analyzing the intersections, we found two solutions from the positive root and two from the negative root.

Adding them together, we get a total of 4 distinct values of $x$. You have navigated the trigonometric forest and emerged victorious!