How do you solve $\frac{cos x cot x}{1 - sin x} = 3$ $(cosxcotx)/(1-sinx)=3$ and find all solutions in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Answer 1 · 2016-09-06T14:05:30

$x = \frac{π}{6} and \frac{5 π}{6}$ $x = pi/6 and (5pi)/6$

$cos x cot x = 3 (1 - sin x)$ $cosxcotx = 3(1 - sinx)$

$cos x cot x = 3 - 3 sin x$ $cosxcotx = 3 - 3sinx$

$cos x \times \frac{cos x}{sin x} = 3 - 3 sin x$ $cosx xx cosx/sinx = 3 - 3sinx$

$\frac{{cos}^{2} x}{sin x} = 3 - 3 sin x$ $cos^2x/sinx = 3 - 3sinx$

${cos}^{2} x = sin x (3 - 3 sin x)$ $cos^2x = sinx(3 - 3sinx)$

${cos}^{2} x = 3 sin x - 3 {sin}^{2} x$ $cos^2x = 3sinx - 3sin^2x$

$1 - {sin}^{2} x = 3 sin x - 3 {sin}^{2} x$ $1 - sin^2x = 3sinx - 3sin^2x$

$3 {sin}^{2} x - {sin}^{2} x - 3 sin x + 1 = 0$ $3sin^2x - sin^2x - 3sinx + 1 = 0$

$2 {sin}^{2} x - 3 sin x + 1 = 0$ $2sin^2x - 3sinx + 1 = 0$

$2 {sin}^{2} x - 2 sin x - sin x + 1 = 0$ $2sin^2x - 2sinx - sinx + 1 = 0$

$2 sin x (sin x - 1) - 1 (sin x - 1) = 0$ $2sinx(sinx - 1) - 1(sinx - 1) = 0$

$(2 sin x - 1) (sin x - 1) = 0$ $(2sinx - 1)(sinx - 1) = 0$

$sin x = \frac{1}{2} and sin x = 1$ $sinx = 1/2 and sinx = 1$

$x = \frac{π}{6}, \frac{5 π}{6} and \frac{π}{2}$ $x = pi/6, (5pi)/6 and pi/2$

However, $\frac{π}{2}$ $pi/2$ is extraneous, since it renders the denominator $0$ $0$ .

Hopefully this helps!

How do you solve (cosxcotx)/(1-sinx)=3cosxcotx1−sinx=3(cosxcotx)/(1-sinx)=3 and find all solutions in the interval [0,2pi)[0,2π)[0,2pi)?