Question

Cos(2theta-pi/2)=-1
Solve the equation on the interval 0_ 12 years ago

Answer 1

It appears that you want to solve for the value of theta, which represents an angle. Assuming that you understand that the inverse cosine function arccos x will "undo" cos(2theta-pi/2) on the left side of the equation: 2theta=pi/2 = arccos(-1).

Answer 2

The right side of this equation could be voiced as "the angle whose cosine equals -1." This angle is pi (or -pi). Thus we end up with theta-pi/2 = pi. Solve for theta.

Answer 3

I want to just get the radical

Answer 4

I don't believe this problem involves a radical, Tim. How so?

Answer 5

Like in this example: 2sin(theta) + 1= 0
You would simplify to get 2sin(theta) = -1,
Then sin(theta) = -1/2
You find what equals sin -1/2 and that is 7pi/2 and 11pi/6, which are the answers.

Answer 6

Tim,what you have found in this example problem are two values for the angle (theta). These values lie within the indicated interval 0<theta<2pi. Again, I see no "radical" here; the angle values 7pi/2 and 11pi/6 are expressed in radians (yes), not as radicals.

Answer 7

I'm sorry, I meant radians, not radicals.

Answer 8

Then you're on track. Please go back to my previous suggestion: Solve theta-pi/2 = pi for theta. Keep theta on the left; move pi/2 to the right. You'll have theta = pi + pi/2, or 3pi/2. This angle, the equivalent of 270 degrees, is within the interval 0<theta<2pi. Hope this makes sense to you.

Answer 9

Thank you.