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how do I solve 4cos(x+4)=1?
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\(\bf 4cos(x+4)=1\implies cos(x+4)=\cfrac{1}{4}\\ \quad \\ \implies cos^{-1}[cos(x+4)]=cos^{-1}\left(\cfrac{1}{4}\right)\)
recall that \(\bf cos^{-1}[cos(\theta)]=\theta\)
\[Cos(x+4) = \frac{ 1 }{ 4 }\] \[x+4 = Cos ^{-1}(1/4)\] \[x = Cos ^{-1}(1/4) -4\] And you need to use the calculator to find the value of arc cos(1/4)
okay thank you that helped a lot!
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