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

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Answer 1 · 2015-04-17T03:33:48

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

$f(x) = (1 + cos x) + sin x = 0$ . 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

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