If cos2x – asinx = 2a – 7, then range of a is

The correct option is 2 ≤ a ≤ 6.

To find the range of the parameter $a$ for which the equation has a real solution, we can simplify the given trigonometric equation:
$\cos \left(2x\right)-a\sin \left(x\right)=2a-7$

First, recall the double-angle identity for cosine in terms of sine:
$\cos \left(2x\right)=1-2{\sin }^2\left(x\right)$

Substitute this identity into the original equation:
$1-2{\sin }^2\left(x\right)-a\sin \left(x\right)=2a-7$

Let us introduce a substitution to simplify the equation. Let:
$t=\sin \left(x\right)$
Since the sine function is bounded, the variable $t$ must satisfy the condition:
$t\in [-1,1]$

Substituting $t$ into the equation gives a quadratic equation in terms of $t$:
$1-2t^2-at=2a-7$
Rearranging the terms:
$2t^2+at+2a-8=0$

We want to find the values of $a$ for which there exists a solution for $t$ in the interval $[-1,1]$. Let us isolate $a$ by grouping the terms containing $a$:
$a(t+2)=8-2t^2$
Factor out a 2 on the right-hand side:
$a(t+2)=2(4-t^2)$
Using the difference of squares factorization:
$a(t+2)=2(2-t)(2+t)$

Since $t\in [-1,1]$, the term $t+2$ lies in the range $[1,3]$ and is never equal to zero. Therefore, we can divide both sides by $(t+2)$:
$a=2(2-t)$

Now, we find the range of values for $a$ as $t$ varies from $-1$ to $1$:
When $t=1$:
$a=2(2-1)=2$
When $t=-1$:
$a=2(2-(-1\left)\right)=2\left(3\right)=6$

Because the function $a\left(t\right)=2(2-t)$ is a continuous linear function, the range of $a$ is bounded by these minimum and maximum values:
$2\le a\le 6$