What is the general solution for (cos(x)-cos^2(x))/(sin(x))=sin(x)cos(x) ?

The general solution for (cos(x)-cos^2(x))/(sin(x))=sin(x)cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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(cos(x)-cos^2(x))/(sin(x))=sin(x)cos(x)

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} = sin (x) cos (x)

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} = sin (x) cos (x)

Subtract

sin (x) cos (x)

from both sides

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} - sin (x) cos (x) = 0

Simplify

\frac{cos ( x ) - cos ^{2} ( x )}{sin ( x )} - sin (x) cos (x) : \frac{cos ( x ) - cos ^{2} ( x ) - sin ^{2} ( x ) cos ( x )}{sin ( x )}

\frac{cos ( x ) - cos ^{2} ( x ) - sin ^{2} ( x ) cos ( x )}{sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (x) - cos^{2} (x) - sin^{2} (x) cos (x) = 0

Factor

cos (x) - cos^{2} (x) - sin^{2} (x) cos (x) : cos (x) (1 - cos (x) - sin^{2} (x))

cos (x) (1 - cos (x) - sin^{2} (x)) = 0

Solving each part separately

cos (x) = 0 or 1 - cos (x) - sin^{2} (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

1 - cos (x) - sin^{2} (x) = 0 : x = 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2 πn

Since the equation is undefined for:

2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

1 = - cot^{2} (x) + 2

60 \cdot \frac{π}{180} = 2 arctan (\frac{x}{17})

sin (a) = - \frac{1}{8}, \frac{3 π}{2} \leq a \leq 2 π

3 cos^{2} (x) - 23 cos (x) = sin^{2} (x)

sin (x) + cos (x) = \frac{3}{2}

What is the general solution for (cos(x)-cos^2(x))/(sin(x))=sin(x)cos(x) ?
The general solution for (cos(x)-cos^2(x))/(sin(x))=sin(x)cos(x) is x= pi/2+2pin,x=(3pi)/2+2pin