Question

How do you solve `[(1+cosx)/(sinx)]=-1` over [0,2pi)?

Answer 1

Here is one way to solve it, by using these trig identities

1. `color(blue)( 1 + cos x = 2*cos^2 (x/2))`
2. `color(blue)(sin x = 2 sin (x/2)*cos (x/2)`
3. `color(blue)(sin x + cos x = sqrt2*sin (x/2 + pi/4)`.

`f(x) = (1 + cos x) + sin x = 0`. Change variable `x` to `x/2`.

`f(x) = 2*cos^2 (x/2) + 2*sin (x/2)*cos (x/2) = 0`

`f(x) = 2cos (x/2)*(cos (x/2) + sin (x/2) = 0`

Next, solve:

`cos (x/2) = 0 --> x/2 = pi/2 and x/2 = 3pi/2 --> x = pi and x = (3pi)/2.`

`(sin x + cos x) = sqrt2*sin (x/2 + pi/4) = 0`

`sin (x/2 + pi/4) = 0 --> x/2 + pi/4 = pi and 2pi`

`x/2 = pi - pi/4 = (3pi)/4 --> x = (6pi)/4 = (3pi)/2`

`x/2 = 2pi - pi/4 --> x/2 = (7pi)/4 -->x = (14pi)/4 = (7pi)/2` (out of interval)
Answers within (`0, 2pi`): `pi`; and `(3pi)/2`

Check:
When `x = pi --> cos x = -1; sin x = 0 --> f(x) = 1 - 1 = 0` Correct
When `x = (3pi)/2 --> cos x = 0; sin x = -1--> f(x) = 1 - 1 = 0` Correct