How do you solve the equation on the interval [0,2pi) for 2sin (t) cos (t) + sin (t) -2 cos (t) -1 =0 ?

1 Answer
Jun 10, 2016

pi/2, (2pi)/3, (4pi)/3

Explanation:

Group the terms of the equation in order to factor it:
f(t) = (2sin t.cos t - 2cos t) + (sin t - 1 )= 0
f(t) = 2cos t(sin t - 1) + (sin t - 1) = 0
f(t) = (sin t - 1)(2cos t + 1) = 0
Solve the 2 binomials by using Trig Table and unit circle -->
a. sin t - 1 = 0 --> sin t = 1 --> t = pi/2
b. 2cos t + 1 = 0 --> cos t = -1/2 --> t = +- (2pi)/3
Note that the arc -(2pi)/3 is co-terminal to arc (4pi)/3
Answer for (0, 2pi)
pi/2, (2pi)/3, (4pi)/3