solve equation with cos and powers

Question

Solve the following equation. $$3^{x^2-6x+11}=8+\cos^2\frac{\pi x}{3}.$$

Here is the problem that I am struggling with it I tried to take a logarithm of both side but I kinda stuck can someone help me with this? Thanks

Answer 1

$$9\geq8+\cos^2\frac{\pi x}{3}=3^{x^2-6x+11}=3^{(x-3)^2+2}\geq9.$$ Thus, $\cos^2\frac{\pi x}{3}=1$ and $x=3$, which gives the answer: $$\{3\}$$

Answer 2

Add 1 to each side of the equation to get

$$3^{x^2-6x+11} + 1 = 9 + \cos^2 \frac {\pi x} {3}$$

Take $\log_3$ of both sides to get

$$x^2-6x+11 = 2 + \log_3 \cos^2 \frac {\pi x} {3}$$

The maximum of $\cos^2 \frac {\pi x} {3} = 1$, thus $\log_3 1 = 0$ , so we have

$$x^2 - 6x + 11 = 2$$

Bringing the 2 to the right hand side we have

$$x^2 - 6x + 9 = 0$$

Factoring gives us

$$(x-3)^2 = 0$$

And thus we have $$x = 3$$

Check: $3^{3^2-18+11} = 3^{9-18+11} = 3^2 = 9; \cos^2 \pi = 0; \Rightarrow 9 + 0 = 9 \ \ \checkmark$