Question

Please help! Precalc/Trig problem :/ Click attached

Answer 1

Ok so what you have to do is manipulate it so that you isolate secx. Once you have done that you know that secant is 1/cosx, so you reaarange for cosx. then it should be easy peasy from there

Answer 2

Oh I see. Can you please tell me if it's the first one? I tried solving it awhile ago but I don't know if I got it right :/

Answer 3

nope you probably calculated something wrong

Answer 4

Hmm :/ Can you go step by step solving it so I can see what I did wrong and fully understand?

Answer 5

Or is it C.?

Answer 6

\[3\sec(x) - 3 = 5 - \sec(x) \implies 4 \sec(x) = 2 \implies \sec(x) = \frac{1}{2} \implies \cos(x) = 2\]

Answer 7

Is that possible??

Answer 8

Sorry calculated wrong..

Answer 9

\[3\sec(x) - 3 = 5 - \sec(x) \implies 4 \sec(x) = 8 \implies \sec(x) = 2 \implies \cos(x) = \frac{1}{2}\]

Answer 10

\[x = 2n \pi \pm \frac{\pi}{3}\]

Put n = 0:

\[x = \frac{\pi}{3}\]


Then put n = 1:
\[x = 2 \pi - \frac{\pi}{3} \implies x = \frac{5\pi}{3}\]

Answer 11

Yes you are right it is C..

Ha ha ha..

Answer 12

Thank you @waterineyes :)

Answer 13

Welcome dear..