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

The general solution for (sec(x)+1)^2-tan(x)=0 is x=pi+2pin,x=-1.97861…+2pin

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(sec(x)+1)^2-tan(x)=0

(sec (x) + 1)^{2} - tan (x) = 0

x = π + 2 πn, x = - 1.97861 \dots + 2 πn

Solution steps

(sec (x) + 1)^{2} - tan (x) = 0

Add

tan (x)

to both sides

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

Square both sides

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

Subtract

tan^{2} (x)

from both sides

(sec^{2} (x) + 2 sec (x) + 1)^{2} - tan^{2} (x) = 0

Rewrite using trig identities

2 + sec^{4} (x) + 4 sec (x) + 4 sec^{3} (x) + 5 sec^{2} (x) = 0

Solve by substitution

sec (x) = - 1, sec (x) \approx - 2.52137 \dots

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 2.52137 \dots : x = arcsec (- 2.52137 \dots) + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

Combine all the solutions

x = π + 2 πn, x = arcsec (- 2.52137 \dots) + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

Verify solutions by plugging them into the original equation

x = π + 2 πn, x = - arcsec (- 2.52137 \dots) + 2 πn

Show solutions in decimal form

x = π + 2 πn, x = - 1.97861 \dots + 2 πn

cos (A) = \frac{8}{17}

sec (θ) = \frac{6}{- 5}, sec (- θ)

cos (π z) = 0

sin (x) \cdot cos (x) - 3 cos (x) = 0

9.8 \cdot sin (α) - 0.2 \cdot 9.8 \cdot cos (α) = 0.47

What is the general solution for (sec(x)+1)^2-tan(x)=0 ?
The general solution for (sec(x)+1)^2-tan(x)=0 is x=pi+2pin,x=-1.97861…+2pin