Question

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

Answer 1

`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`