Question

I need some help setting this problem up to find all the solutions to this equation.

sec(θ/2) = cos (θ/2)

Answer 1

Do I use the half angle formula? And if so, how do I handle the sec(θ/2)? I know sec is 1/cos but I'm not sure how to factor that into the formula...

Answer 2

since sec(theta) = 1/cos(theta)

we go from this

sec(θ/2) = cos (θ/2)

to this

1/cos(θ/2) = cos (θ/2)

Answer 3

then you multiply both sides by cos(θ/2)

1/cos(θ/2) = cos (θ/2)

1 = cos (θ/2)*cos(θ/2)

1 = cos^2 (θ/2)

cos^2 (θ/2) = 1

do you see what to do next?

Answer 4

When I multiply both sides by cos(θ/2) I get

θ/2 = cos^2(θ/2) ?

Answer 5

think of cos(θ/2) as x

Answer 6

so 1/cos(θ/2) = cos (θ/2) is really the same as 1/x = x

Answer 7

Ok, let me write this out again on my paper...

Answer 8

\(\bf sec\left(\frac{\theta}{2}\right)=cos\left(\frac{\theta}{2}\right)\implies \cfrac{1}{cos\left(\frac{\theta}{2}\right)}=cos\left(\frac{\theta}{2}\right)
\\ \quad \\
1=cos^2\left(\frac{\theta}{2}\right)\implies \sqrt{1}=\sqrt{cos^2\left(\frac{\theta}{2}\right)}\implies 1=cos\left(\frac{\theta}{2}\right)
\\ \quad \\
1=\pm\sqrt{\cfrac{1+cos(\theta)}{2}}\implies 1^2=\pm\left[\sqrt{\cfrac{1+cos(\theta)}{2}}\right]^2\implies 1=\pm\cfrac{1+cos(\theta)}{2}\)

Answer 9

as pointed out by jim_thompson5910 , so you'd end up using the half-angle identity

Answer 10

The way you wrote out 1/cos(θ/2) was helpful. I know it's the same thing, but seeing it written that way makes it make sense.

Answer 11

I know I'm slow, but I think I figured it out.
cos(θ) must then equal 1, which is 2π. So my answers would then be 2π + 2πk.

Answer 12

and you can simplify that to say x = 2πk where k is any integer

Answer 13

Thanks!

Answer 14

you're welcome