Animated Solution for Mathematics - Trigonometry: Find the values of $x\in (-\pi ,\pi )$ which satisfy the equation $8^{(1+|\cos x|+{\cos }^{2}x+|{\cos }^{3}x|+\dots )}=4^3$.

When you first look at the equation $8^{(1+|\cos x|+{\cos }^{2}x+|{\cos }^{3}x|+\dots )}=4^3$, the infinite series in the exponent may seem daunting. However, we can simplify this by observing the pattern of the terms.

We know that ${\cos }^{2}x=|\cos x{|}^2$ and $|{\cos }^{3}x|=|\cos x{|}^3$. Substituting these into the exponent, we identify an infinite Geometric Progression (G.P.) with the first term $a=1$ and a common ratio $r=|\cos x|$.

The sum of an infinite G.P. is given by the formula $S=\frac{a}{1-r}$. Substituting our values, we obtain:

Now, we substitute the sum $S$ back into the original equation: $8^S=4^3$. To solve this, we express both sides using a common base of $2$.

Since $8=2^3$ and $4=2^2$, the equation becomes:

Applying the power of a power rule, we get $2^{3S}=2^6$. Equating the exponents, we find $3S=6$, which simplifies to $S=2$.

We now equate our expression for $S$ to the value we just calculated:

Cross-multiplying gives $1=2(1-|\cos x|)$, which simplifies to $1=2-2|\cos x|$. Rearranging the terms, we find:

For $x\in (-\pi ,\pi )$, the absolute value implies $\cos x=\frac{1}{2}$ or $\cos x=-\frac{1}{2}$.

1. For $\cos x=\frac{1}{2}$, the solutions are $x=\frac{\pi }{3}$ and $x=-\frac{\pi }{3}$. 2. For $\cos x=-\frac{1}{2}$, the solutions are $x=\frac{2\pi }{3}$ and $x=-\frac{2\pi }{3}$.

The final set of solutions is $x\in \{-\frac{2\pi }{3},-\frac{\pi }{3},\frac{\pi }{3},\frac{2\pi }{3}\}$.