Question

I have to find all the solutions between the interval (0,2pi), so basically all possible solutions on the unit circle and the problem is as follows: sec(x+pi/3)+(sqrt 2)=0. I know there is are identities such as sin(a+b), and I was wondering if there was one for sec. If not, how would I go about solving this? Thank you!

Answer 1

try \[\frac{1}{\cos(x+\pi/3)}=-\sqrt{2}\]

Answer 2

or possibly by cross multiplication, we can have \[\cos(x+\pi/3)=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\]knowing that \[\cos^{-1}(-\sqrt{2}/2)=3\pi/4~and~5\pi/4\]we can solve for x...

Answer 3

what do you think @tholm001 ?

Answer 4

from \[\cos (x+\pi/3)=-\sqrt{2}/2\]we can have \[x+\pi/3=\cos^{-1}(-\sqrt{2}/2)\]or\[x+\pi/3=5\pi/4\]\[x=(5\pi/4)-(\pi/3)=(15\pi-4\pi)/12=11\pi/12\]and if\[x+\pi/3=3\pi/4\]\[x=(3\pi/4)-(\pi/3)=(9\pi-4\pi)/12=5\pi/12\]

Answer 5

isn't sec=1/sin. So Amplitude=1 B=1(period 2pi/b) Shift left=Pi/3 VT= sqrt 2 0<x<pi/2. So pi/3=60deg or
sin60=sqrt3/2 ---csc is reciprocal
cos60=1/2 ---sec rec
tan60=3/1 cot 1/3

Answer 6

from definition\[\sec x=1/\cos x\]

Answer 7

in the interval \((0,2\pi)\), the solution is that \(x=5\pi/12\) or \(x=11\pi/12\).

Answer 8

any questions @tholm001 ?

Answer 9

This makes it so much clearer for me! thank you so much! you are so helpful.