Question

i need help for a math problem which is pre calc and how i would simplify it....it is sec^5(x)=4sec(x)

Answer 1

\[\sec(x) = \frac{ 1 }{ \cos(x) }\] and \[\sec^2(x) = 1-\tan^2(x)\]

Answer 2

Using the first rule that abbot mentioned,\[\large \sec^5x=4\sec x \qquad \rightarrow \qquad \frac{1}{\cos^5x}=4\frac{1}{\cos x}\]Doing some cross multiplication gives us,\[\large \cos x=4 \cos^5x\]\[\large 4\cos^5x-\cos x=0\]
Factoring out a cos x from each term,\[\large \cos x(4\cos^4x-1)=0\]

Can you figure out what to do next? :)

Answer 3

Yes i think so, haha i think i was over thinking things thanks ill just comebsack if i still don't get it.