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

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

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sec(x)+1=tan(x)

sec (x) + 1 = tan (x)

x = 2 πn + π

Solution steps

sec (x) + 1 = tan (x)

Subtract

tan (x)

from both sides

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

Express with sin, cos

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

Simplify

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

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

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

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

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

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

Since the equation is undefined for:

\frac{π}{2} + 2 πn

x = 2 πn + π, x = π + 2 πn

Merge Overlapping Intervals

x = 2 πn + π

2 sin^{2} (x) + 5 sin (x) + 2 = 0

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

tan^{2} (x) + tan (x) - 2 = 0

0 = 1 + cos (x)

8 sin^{2} (θ) + 17 sin (θ) + 9 = 0

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