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

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

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

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

x = 0, x = 2 π

Solution steps

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

Subtract

1

from both sides

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

Express with sin, cos

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

Simplify

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

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

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

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

Add

cos (x)

to both sides

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

Square both sides

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

Subtract

cos^{2} (x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = 1, sin (x) = 0

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

Combine all the solutions

x = \frac{π}{2}, x = 0, x = π, x = 2 π

Verify solutions by plugging them into the original equation

x = 0, x = 2 π

3 tan (\frac{x}{9}) - 3 = 0

x, tan (x) = \frac{1}{3}

cos (x) = \frac{9}{12}

2 tan^{2} (x) + 1 = sec^{2} (x) + sin^{2} (x) + cos^{2} (x)

cos (x) = - \frac{1}{6}

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