What is the general solution for 1-tan(x)=sec(x) ?

The general solution for 1-tan(x)=sec(x) is x=2pin,x=2pi+2pin

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1-tan(x)=sec(x)

1 - tan (x) = sec (x)

x = 2 πn, x = 2 π + 2 πn

Solution steps

1 - tan (x) = sec (x)

Subtract

sec (x)

from both sides

1 - tan (x) - sec (x) = 0

Express with sin, cos

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

Simplify

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

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

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

cos (x) - sin (x) - 1 = 0

Add

sin (x)

to both sides

cos (x) - 1 = sin (x)

Square both sides

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

Subtract

sin^{2} (x)

from both sides

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

Factor

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

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

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

x = 2 πn, x = 2 π + 2 πn

sec (\frac{3 θ}{2}) = - 2, 0 \leq θ < 2 π

tan (x - \frac{π}{6}) = 3

- 5 cos (5 x) = 0

2 sin (x) tan (x) + 5 = 4 sec (x)

cos^{2} (x) - cos (x) - 2 = 0, 0 \leq x \leq 2 π

What is the general solution for 1-tan(x)=sec(x) ?
The general solution for 1-tan(x)=sec(x) is x=2pin,x=2pi+2pin