cos squared of theda - cos of theda equals 0?

Answer 1

cos^2x-cosx=0
cosx(cosx-1)=0
cosx=0 or x=(2n+1)pi/2
cosx=1 or x=2npi

Answer 2

Instead of theta, I'll use x. So you want to solve:

cos^2(x) - cos(x) = 0

Your first step would be to factor this. Pull out a cos(x) from each term, to turn it into a product.

cos(x) [cos(x) - 1] = 0

Now, equate each factor to 0.

cos(x) = 0
cos(x) - 1 = 0 ------> cos(x) = 1

Individually solve both of those equations. Let's start with
cos(x) = 0.

Where is cos(x) equal to 0? Assume x is restricted from 0 to 2pi, this happens when x = pi/2 or x = 3pi/2.

Where is cos(x) = 1? This happens at x = 0.

Therefore, x = {pi/2, 3pi/2, 0}

If, however, we wanted the general solution where x is NOT restricted from 0 to 2pi, we just add 2k(pi) to each solution.

x = {pi/2 + 2k(pi) | k an integer} U
{3pi/2 + 2k(pi) | k an integer} U
{0 + 2k(pi) | k an integer}

Answer 3

cos^2 theta - cos theta = 0
<=> cos theta = 0 or cos theta = 1
<=> theta = pi/2 + k*pi or theta = k*2pi

Answer 4

cos^2 th-cos th=cos th(cos th -1) equals zero only when cos th=1, th=0, 2π, 4π, 6V etc
or cos th=0
th=π/2, 3π/2, 5π/2 etc