What is the general solution for -1-tan(x)=sec(x),0<= x<2pi ?

The general solution for -1-tan(x)=sec(x),0<= x<2pi is x=pi

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-1-tan(x)=sec(x),0<= x<2pi

- 1 - tan (x) = sec (x), 0 \leq x < 2 π

x = π

Solution steps

- 1 - tan (x) = sec (x), 0 \leq x < 2 π

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = - 1, 0 \leq x < 2 π : x = π

Combine all the solutions

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

Verify solutions by plugging them into the original equation

x = \frac{3 π}{2}, x = π

Since the equation is undefined for:

\frac{3 π}{2}

x = π

cosh (z) + 1 = 0

tan (x) = 11, sin (x)

4 sin (x) = 2.54648

y, \frac{3}{2} = sin (arcsin (0) + y)

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

What is the general solution for -1-tan(x)=sec(x),0<= x<2pi ?
The general solution for -1-tan(x)=sec(x),0<= x<2pi is x=pi