What is the general solution for (sin(x))/(1-cos^2(x))=sec(x) ?

The general solution for (sin(x))/(1-cos^2(x))=sec(x) is x= pi/4+pin

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(sin(x))/(1-cos^2(x))=sec(x)

\frac{sin ( x )}{1 - cos ^{2} ( x )} = sec (x)

x = \frac{π}{4} + πn

Solution steps

\frac{sin ( x )}{1 - cos ^{2} ( x )} = sec (x)

Subtract

sec (x)

from both sides

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

Simplify

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

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

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

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

cos (2 x) - 2 sin (x) = 1, 0^{\circ} < x \leq 36 0^{\circ}

cos (aπ) - 1 = 0

3 tan (2 θ) - 3 = 0

cos (θ) = \frac{4 k - 1}{3}

5 sin (x) - 4 = 0

What is the general solution for (sin(x))/(1-cos^2(x))=sec(x) ?
The general solution for (sin(x))/(1-cos^2(x))=sec(x) is x= pi/4+pin