VicKmath7 wrote: Find all real $a$ such that the equation $3^{\cos (2x)+1}-(a-5)3^{\cos^2(2x)}=7$ has a real root. So we are looking for the range of $a=5+\frac{3^{1+\cos 2x}-7}{3^{\cos^22x}}$ over $\mathbb R$ And so the range of $f(x)=5+\frac{3^{1+x}-7}{3^{x^2}}$ over $[-1,+1]$ $f(x)=5+3^{1+x-x^2}-7.3^{-x^2}$ and so $f'(x)=(\ln 3) 3^{-x^2}\left((1-2x)3^{x+1}+14x\right)$ It is easy to see that $(1-2x)3^{x+1}+14x$ over $[-1,1]$ has a unique zero $x_0\in\left(-\frac 14,-\frac 15\right)$ and is negative before $x_0$ and positive after. So $f(x)$ is decreasing over $[-1,x_0]$ from $3\to f(x_0)$ and increasing over $[x_0,+1]$ from $f(x_0)\to \frac{17}3$ And so answer is $a\in\left[f(x_0),\frac{17}3\right]$ I did not find any smart closed form for $x_0$ and $f(x_0)$ and easy numerical calculus gives : $x_0\sim -0.2438737752153$ and $f(x_0)\sim 0.5925003438768$ And answer $\boxed{a\in\left[\sim 0.5925003438768,\frac{17}3\right]}$