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

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

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

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

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= 2 tan^{2} (x)

2 tan^{2} (x) = 0

Divide both sides by

2

2 tan^{2} (x) = 0

Divide both sides by

2

2 tan^{2} (x) = 0

Divide both sides by

2

\frac{2 tan ^{2} ( x )}{2} = \frac{0}{2}

Simplify

tan^{2} (x) = 0

tan^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

tan (x) = 0

General solutions for

tan (x) = 0

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 0 + πn

x = 0 + πn

Solve

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan^{2} (x) - 3 tan (x) = 0, 0 \leq x < 2 π

tan (θ) = 10

8 sin (θ) - 5 = 3 sin (θ)

2 cos (4 x) + 3 = 0

cos (θ) = 0.4

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