Answer:
x=0, x=1.772
Explanation:
Given the equation:
[tex]-5\cos ^2(x)+4\cos (x)+1=0[/tex]
Let cos(x)=p
[tex]\begin{gathered} -5\cos ^2(x)+4\cos (x)+1=0 \\ \implies-5p^2+4p+1=0 \end{gathered}[/tex]
First, solve the equation above for p:
[tex]\begin{gathered} -5p^2+5p-p+1=0 \\ -5p(p-1)-1(p-1)=0 \\ (-5p-1)(p-1)=0 \\ -5p-1=0\text{ or }p-1=0 \\ \implies p=-\frac{1}{5}\text{ or }p=1 \end{gathered}[/tex]
Recall that we made the substitution: cos(x)=p
When p=1
[tex]\begin{gathered} \cos (x)=1 \\ x=\cos ^{-1}(1) \\ x=0 \end{gathered}[/tex]
When p=-1/5
[tex]\begin{gathered} \cos (x)=-\frac{1}{5} \\ x=\cos ^{-1}(-\frac{1}{5}) \\ x=1.772\text{ (in radians)} \end{gathered}[/tex]
The smallest non-negative solutions to the equation are 0 and 1.772.
