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

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

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

tan^{2} (x) = sec (x) - 1, 0 \leq x \leq 2 π

x = 0, x = 2 π

Solution steps

tan^{2} (x) = sec (x) - 1, 0 \leq x \leq 2 π

Subtract

sec (x) - 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sec (x) = 1, sec (x) = 0

sec (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

sec (x) = 0, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = 0, x = 2 π

sin^{2} (θ) = 2 sin^{2} (\frac{θ}{2})

cos (x) = cos (x)

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

2 cos^{3} (x) + cos^{2} (x) - cos (x) = 0

0.26 = sin (0.25 π t)

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