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

The general solution for (sec(x))/(tan(x))=(cos(x))/((sin^2(x))) is x= pi/4+pin

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(sec(x))/(tan(x))=(cos(x))/((sin^2(x)))

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

x = \frac{π}{4} + πn

Solution steps

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

Subtract

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

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

- sin (x) + sec (x) sin^{2} (x) = 0

Factor

- sin (x) + sec (x) sin^{2} (x) : sin (x) (- 1 + sec (x) sin (x))

sin (x) (- 1 + sec (x) sin (x)) = 0

Solving each part separately

sin (x) = 0 or - 1 + sec (x) sin (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

- 1 + sec (x) sin (x) = 0 : x = \frac{π}{4} + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

Since the equation is undefined for:

2 πn, π + 2 πn

x = \frac{π}{4} + πn

x, tan (x) = \frac{1}{8}

sin (x) + 7 = 7

cos (3 x) = 0, 0 < x < π

12 csc (x) = 3 csc (x) + 63

2 - 2 csc^{2} (x) = - csc^{2} (x) + csc (x)

What is the general solution for (sec(x))/(tan(x))=(cos(x))/((sin^2(x))) ?
The general solution for (sec(x))/(tan(x))=(cos(x))/((sin^2(x))) is x= pi/4+pin