gobathegreat wrote: Find all values of real parameter $a$ for which equation $2{\sin}^4(x)+{\cos}^4(x)=a$ has real solutions Easy to see that $f'(x)$ is zero at $k\frac{\pi}2$ and when $cos^2x=\frac 23$ And at these points $f(x)$ is $1,2$ or $\frac 23$ Hence the answer $\boxed{a\in\left[\frac 23,2\right]}$

What is wrong in this solution? If $sin^2x=t$, and we use $cos^4x=(1-sin^2x)^2=t^2-2t+1$, so we get this equation: $3t^2-2t+1-a=0$, and this must have real, positive solutions, so $1-a\ge 0$, and $\Delta =4a\ge 0\Rightarrow a\in \left[ 0,1 \right]$

Let $f:[0,1]\to\mathbb{R}, f(t)=3t^2-2t+1$. The minimum of the function $f$ occurs for $t_m=\dfrac{1}{3}$ and $\min{f(t)}=f(t_m)=\dfrac{2}{3}$. The function $f$ is strictly decreasing on $[0,t_m]$ and is strictly increasing on $[t_m,1]$. Results: $\max{f(t)}=\max{(f(0), f(1))}=f(1)=2$. Hence, $a\in\left[\dfrac{2}{3}, 2\right]$. Return to TuZo's solution: The equation $3t^2-2t+1-a=0$ must have at least a solution $t_1$ in $[0,1]$ (not necessarily both) and $\Delta=4(3a-2)$ (not $4a$). The conditions for the root $t_1\in[0,1]$ are: $\Delta\ge0$ and $f(0)\cdot f(1)\le0$ or $\left(f(0)\ge0, f(1)\ge0, t_m=\dfrac{1}{3}\in[0,1]\right)$.

OK, thank you, it was a banal error in my delta.

gobathegreat wrote: Find all values of real parameter $a$ for which equation $2{\sin}^4(x)+{\cos}^4(x)=a$ has real solutions Other method : 1) $2\sin^4x+\cos^4x=\frac 23+3(\sin^2x-\frac 13)^2\ge\frac 23$ (reached when $\sin^2=\frac 13$) 2) $2\sin^4x+\cos^4x=2-\cos^2x(3\sin^2x+1)\le 2$ (reached when $\cos x=0$)