Trigonometry question??

Question

if i had to find o(theta) in radians how would i go about doing this for

seco=2coso

thanks i have alot of questions like this but i think if this one is explained to me i will be able to do it

Answer 1

sec  t = 1/cost so 1/cos t = 2 cos t and cos^2t =1/2 and

cos t = +-1/2 sqrt2

so t = pi/4 +2kpi and t = 7pi/4+2kpi

also t= 3pi/4+2kpi and t= 5pi/4+2kpi ( k any integer

Answer 2

Alrighty, lets see here.

we know that sec(x) (i'm using x for thetha) is the same thing as 1/cos(x)

So our equation becomes:

(1/cos (x)) = 2cos(x)  multiply both sides by cos(x)

1 = 2cos^2(x)  divide both sides by 2

1/2 = cos^2(x)  square root both sides

+/- sqrt(2)/2 = cos(x)  now using the unit circle, we know that cos(x) = +/- sqrt(2)/2 at many angles

x = pi/4, 3pi/4, 5pi/4, 7pi/4

if no set domain is given, you can write your answer in the form:

x = pi/4 + k(pi/2), where k is any integer (this will give all the solutions since trig functions are periodic and therefore continue in the same pattern over and over again)

voila

Answer 3

well we know seco= 1/cos0

1/cos0 = 2cos0

multiply both side by cos0

1= 2 cos^2(0)

divide by 2

1/2 = cos^2(0)

take square root both sides

+/- sqrt(1/2) = cos0

+/- 1/sqrt(2) = cos 0

if cos0= 1/sqrt(2)

so 0=  45 degree or pi/4 in radian

if cos0= -1/sqrt(2)

so 0 = 135 degree or  3pi/4

assume that 0 from 0< theta< 2pi