Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $4{\sin }^{2}x-4{\cos }^{3}x+9-4\cos x=0$; $x\in [-2\pi ,2\pi ]$ is :

The given trigonometric equation is:

This equation is currently difficult to solve because it involves both $\sin x$ and $\cos x$. To simplify, we use the fundamental identity ${\sin }^{2}x=1-{\cos }^{2}x$ to express the entire equation in terms of $\cos x$.

Substituting the identity into the equation, we obtain:

Expanding the terms yields:

Combining the constants $4$ and $9$, we get:

Multiplying by $-1$ and rearranging in descending powers of $\cos x$, we arrive at the cubic equation:

Letting $t=\cos x$, we define the function:

We must remember the golden rule of trigonometry: for any real $x$, the variable $t$ must be strictly bounded within the interval $t\in [-1,1]$.

To determine if $f(t)$ has any roots in the interval $[-1,1]$, we analyze its behavior using calculus. We differentiate $f(t)$ with respect to $t$:

Factoring out a $4$, we have:

The discriminant of the quadratic $3t^2+2t+1$ is $D=2^2-4(3)(1)=4-12=-8$. Since the discriminant is negative and the leading coefficient is positive, $f^{\prime }(t)>0$ for all real $t$.

This implies that $f(t)$ is a strictly increasing function.

Because $f(t)$ is strictly increasing, it can cross the x-axis at most once. To check if it crosses the axis within our valid domain, we evaluate the function at the rightmost endpoint, $t=1$:

Since the maximum value of $f(t)$ on the interval $[-1,1]$ is $-1$, which is less than zero, the function $f(t)$ never reaches zero within the valid domain.

Consequently, there are no values of $t$ that satisfy the equation, and therefore no real values of $x$ exist. The total number of solutions is 0.