Solve a Trig Equation Containing secant

[nickb145]

Homework Statement

sec(5x)-5=0

Homework Equations

The Attempt at a Solution

I turned it into 1/cos(5x)=5 x=θ/5
then i switched it to cos(5x)=/1/5

Then cos^-1(1/5)=θ which = 1.36 (one solution) then divided by 5 which is .2738 which works as one solution
since the positive cos is in the 1st and 4th quadrant i need to subtract 2pi to 1.36= 4.91 then divide by 5 which gives me .9827 which also works

But i am stuck after that

 

[Mark44]

Homework Statement

sec(5x)-5=0

Homework Equations

The Attempt at a Solution

I turned it into 1/cos(5x)=5 x=θ/5
then i switched it to cos(5x)/1/5

And lost your equation.

You're making this harder than it needs to be. If 1/cos(5x) = 5, then cos(5x) = 1/5.

You could also say that cos(5x + k*2$\pi$) = 1/5, and since cosine is positive in Q IV, you could also say that cos(-5x + k*2$\pi$) = 1/5, where k is an integer.

Now you can take cos^-1 of both sides.

Then cos^-1(1/5)=θ which = 1.36 (one solution) then divided by 5 which is .2738 which works as one solution
since the positive cos is in the 1st and 4th quadrant i need to subtract 2pi to 1.36= 4.91 then divide by 5 which gives me .9827 which also works

But i am stuck after that