How do you solve #2cos^2 theta - cos theta -1 = 0#?

As you can see, this is a quadratic equation in cosine. It can be solved by factoring.

As you probably know, trinomials of the form #y = ax^2 + bx + c, a != 1# can be solved by finding two numbers that multiply to #a xx c# and that add to #b#.

#:.# Two numbers that multiply to #-2# and that add to #-1# are #-2# and #+1#.

#(2cos^2theta - 2costheta) + (costheta - 1) = 0#

Factor out a common factor from both pairs of binomials (encased in the parentheses). Note that when you're fully competent at this task the parentheses are unnecessary. For this case, they were just used to focus your attention.

#2costheta(costheta - 1) + 1(costheta - 1) = 0#

#(2costheta + 1)(costheta - 1) = 0#

#costheta = -1/2 and costheta = 1#

By special angles:

#theta = 120˚, 240˚ and 0˚, or (3pi)/2 + 2nπ, (4pi)/3 + 2nπ and 2nπ#, n being an integer.

Hopefully this helps!

Solve the quadratic equation for cos t.
#f(t) = 2cos^2 t - cos t - 1 = 0#
Since a + b + c = 0, use shortcut. The 2 real roots are:
cos t = 1 and #cos t = c/a = - 1/2#
Trig table and unit circle -->

a. cos t = 1 --> arc #t = 0# and #t = 2pi.#
b. #cos t = - 1/2#. Two solution arcs:
#t = +- (2pi)/3#
The arc #(-2pi)/3# is co-terminal to arc #(4pi)/3#.
Answers for (0, 2pi):
#0, (2pi)/3, (4pi)/3, 2pi#