Respuesta :
Step 1
Subtract 4cosx from both sides
[tex]4\text{sin}2\text{x.sinx}-4\cos x=4\cos x-4\cos x[/tex]Simplify
[tex]4\text{sin}2\text{x.sinx}-4\cos x=0[/tex]Step 2
Rewrite using trigonometric identities
[tex]\begin{gathered} \sin 2x=2sinx.\cos x \\ 4(2\sin x\text{cosx})(\sin x)-4\cos x=0 \\ 8\sin ^2(x)\cos (x)-4\cos (x)=0 \end{gathered}[/tex]Step 3
Factorize
[tex]8\sin ^2(x)\cos (x)-4\cos (x)=4\cos (x)(\sqrt[]{2}\sin (x)+1)(\sqrt[]{2}\sin (x)-1)_{}[/tex][tex]\begin{gathered} 4\cos \mleft(x\mright)\mleft(\sqrt{2}\sin \mleft(x\mright)+1\mright)\mleft(\sqrt{2}\sin \mleft(x\mright)-1\mright)=0 \\ \end{gathered}[/tex]Solve separately for x
[tex]\begin{gathered} 4\cos x=0 \\ x=\cos ^{-1}(0) \\ x=90\text{ and }270\text{ within the range \lbrack{}0,2}\pi) \\ In\text{ radians }90^o\text{ = }\frac{\pi}{2},270^o=\frac{3\pi}{2} \\ \\ \end{gathered}[/tex][tex]undefined[/tex]