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Can someone check if my process is correct for this trigonometric question?
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\[-2+\sec3\theta = 0\]
then \[\frac{ 1 }{ \cos 3\theta } -2=0\]
next \[1 = 2 \cos 3\theta\]
then \[\frac{ 1 }{ 2 } = \cos 3\theta\]
so far so good
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okay. So then next do I go \[3\theta = 30+360k\] because the cos of 60 = 1/2
\[3\theta = 60*+360k\]
and then \[\theta = 20 +120k\]
yes that's it and it can also be 20 - 120k
Thanks!
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jojo, don't forget about the other set of solutions. cosine gives us a positive 1/2 in quadrant 4 as well, ya? \[\large\rm 3\theta=300+360k\]\[\large\rm \theta=100+120k\]
Ah yes. I'm using the unit circle link you gave me. xP
oh hehe
Thanks for reminding me. :)
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