MedVision ad

Easy Trig (1 Viewer)

sinophile

Well-Known Member
Joined
Oct 25, 2008
Messages
1,339
Gender
Undisclosed
HSC
N/A
Solve for 0<theta<360:

My solution:
sec^2(theta)=4/3
1/cos^2(theta)=4/3
cos^2(theta)=3/4
cos(theta)=plus-or-minus sqrt(3/4)

Related angle= cos^-1(plus-or-minus 3/4)
approx. =41 degrees
Therefore theta= 41, 139, 221, 319.

Textbook solution:
theta=30, 150, 210, 330.

Why is my solution different from the textbook one? Did I do something wrong?
 

clintmyster

Prophet 9 FTW
Joined
Nov 12, 2007
Messages
1,067
Gender
Male
HSC
2009
Uni Grad
2015
sinophile said:
Solve for 0<theta><360:

My solution:
sec^2(theta)=4/3
1/cos^2(theta)=4/3
cos^2(theta)=3/4
cos(theta)=plus-or-minus sqrt(3/4)

Related angle= cos^-1(plus-or-minus 3/4)
approx. =41 degrees
Therefore theta= 41, 139, 221, 319.

Textbook solution:
theta=30, 150, 210, 330.

Why is my solution different from the textbook one? Did I do something wrong?
maybe you didnt sqrroot the whole thing

your solutions should be cos^-1 (+- root3/2)
</theta>
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
sinophile said:
Solve for 0<?xml:namespace prefix = theta<360 /><theta<360:< p>
My solution:
sec^2(theta)=4/3
1/cos^2(theta)=4/3
cos^2(theta)=3/4
cos(theta)=plus-or-minus sqrt(3/4)

Related angle= cos^-1(plus-or-minus 3/4)
approx. =41 degrees
Therefore theta= 41, 139, 221, 319.

Textbook solution:
theta=30, 150, 210, 330 within given domain

presumable 0 <= x <= 360

Why is my solution different from the textbook one? Did I do something wrong?
sec^(@) = 4/3
cos^2(@) = 3/4
cos @= +- sqrt(3)/2
@ = 30, 150, 210 and 330

</theta<360:<>
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top