We can rewrite the equation as:
$$8{\sin }^2(5x)+\frac{1}{2}\sin (10x)+{\cos }^2(5x)=4$$

Using the double angle formula for sin$2x$ and cos$2x$:
$$8{\sin }^2(5x)+2\sin (5x)\cos (5x)+{\cos }^2(5x)=4$$

Since ${\sin }^2(x)+{\cos }^2(x)=1$:
$$8{\sin }^2(5x)+2\sin (5x)\cos (5x)+1=4$$

Let $u=\sin (5x)$:
$$8u^2+2u+1=4$$ $$8u^2+2u-3=0$$

Solving this quadratic equation by factoring or using the quadratic formula will give the solutions for $u$, which can then be substituted back in for $sin(5x)$ to solve for $x$.