[tex]Let\ \dfrac{x}{2}=a,\ therefore\ x=2a.\\\\\cos^2x+4\sin^2\dfrac{x}{2}=\cos^22a+4\sin^2a\\\\\text{use}\ \cos2x=\sin^2x-\cos^2x\\\\=(\sin^2a-\cos^2a)^2+4\sin^2a\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=(\sin^2a)^2-2(\sin^2a)(\cos^2a)+(\cos^2a)^2+4\sin^2a\\\\=\sin^4a-2\sin^2a\cos^2a+\cos^4a+4\sin^2a\\\\=\underbrace{\sin^4a+2\sin^2a\cos^2a+\cos^4a}_{(*)}-4\sin^2a\cos^2a+4\sin^2a\\\\\text{use}\ (*)\qquad(a+b)^2=a^2+2ab+b^2[/tex]
[tex]=\underbrace{(\sin^2a)^2+2\sin^2a\cos^2a+(\cos^2a)^2}_{(*)}-4\sin^2a(\cos^2a-1)\\\\=(\sin^2a+\cos^2a)^2-4\sin^2a(\cos^2a-1)\\\\\text{use}\ \sin^2a+\cos^2a=1\to\sin^2a=\cos^2a-1\\\\=1^2-4\sin^2a(\sin^2a)=1-4\sin^4a=1-(2\sin^2a)^2[/tex]
[tex]\cos^22a+4\sin^2a=1\\\\1-(2\sin^2a)^2=1\qquad\text{subtract 1 from both sides}\\\\-(2\sin^2a)^2=0\to2\sin^2a=0\qquad\text{divide both sides by 2}\\\\\sin^2a=0\to\sin a=0\\\\a=k\pi\ for\ k\in\mathbb{Z}\\\\\dfrac{x}{2}=k\pi\qquad\text{multiply both sides by 2}\\\\\boxed{x=2k\pi\ for\ k\in\mathbb{Z}}[/tex]
