What is the general solution for tan(x)+sec(x)=0 ?

The general solution for tan(x)+sec(x)=0 is No Solution for x\in\mathbb{R}

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tan(x)+sec(x)=0

tan (x) + sec (x) = 0

No Solution for x \in R

Solution steps

tan (x) + sec (x) = 0

Rewrite using trig identities

sec (x) + tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= sec (x) + \frac{sin ( x )}{cos ( x )}

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

= \frac{sin ( x )}{\frac{1}{s e c ( x )}} + sec (x)

\frac{sin ( x )}{\frac{1}{s e c ( x )}} = sin (x) sec (x)

\frac{sin ( x )}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{sin ( x ) sec ( x )}{1}

Apply rule

\frac{a}{1} = a

= sin (x) sec (x)

= sin (x) sec (x) + sec (x)

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

Factor

sec (x) + sec (x) sin (x) : sec (x) (sin (x) + 1)

sec (x) + sec (x) sin (x)

Factor out common term

sec (x)

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

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

Solving each part separately

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

sec (x) = 0 :

No Solution

sec (x) = 0

sec (x) \leq - 1 or sec (x) \geq 1

No Solution

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

sin (x) + 1 = 0

Move

1

to the right side

sin (x) + 1 = 0

Subtract

1

from both sides

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

Simplify

sin (x) = - 1

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

No Solution for x \in R

cos (θ) = \frac{3}{2}

cos^{2} (θ) + 2 sin (θ) cos (θ) - sin^{2} (θ) = 0

sin (x) - cos (x) = 0, 0 \leq x \leq 2 π

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

\frac{tan ( x )}{2} = 1

What is the general solution for tan(x)+sec(x)=0 ?
The general solution for tan(x)+sec(x)=0 is No Solution for x\in\mathbb{R}