Respuesta :
Answer:
The answer is "2nπ".
Step-by-step explanation:
Given:
[tex]4 \cos x= -\sin^2x+4.......(1)[/tex]
We know:
[tex]\Rightarrow \sin^2 x+\cos^2 x=1\\\\\Rightarrow \sin^2 x= 1 -\cos^2 x\\[/tex]
put the value of [tex]\sin^2 x[/tex] value in the above equation:
[tex]\Rightarrow 4 \cos x= - (1-\cos^2 x)+4\\\\\Rightarrow 4 \cos x= - 1+\cos^2 x+4\\\\\Rightarrow 4 \cos x= \cos^2 x+3\\\\\Rightarrow \cos^2 x-4 \cos x+3=0\\\\[/tex]
Let [tex]\cos x= A[/tex]
[tex]\Rightarrow A^2-4A+3=0 \\ \Rightarrow A^2-(3A+A)+3=0 \\\Rightarrow A^2-3A-A+3=0\\\Rightarrow A(A-3)-1(A-3)=0\\\Rightarrow (A-3)(A-1)=0 \\[/tex]
[tex]\Rightarrow A- 3=0 \ \ \ \ \ \ \ \ \ \ \ \Rightarrow A -1 =0 \\\\[/tex]
[tex]\Rightarrow A= 3\ \ \ \ \ \ \ \ \ \ \ \Rightarrow A =1 \\\\\Rightarrow \cos x = 3\ \ \ \ \ \ \ \Rightarrow \cos x =1\\\\\Rightarrow x = \cos^{-1} 3\ \ \ \ \ \ \ \Rightarrow \cos x =\cos 0\\\\\Rightarrow x = \cos^{-1} 3\ \ \ \ \ \ \ \Rightarrow x = 0\\\\[/tex]
The value of x is [tex]2n\pi\ \ \ _{where} \ \ \ \ \ \ \ n=1, 2, 3......[/tex]
[tex]\boxed{\bold{x=2 n \pi}}[/tex]
