What is the general solution for sin^2(x)-cos^2(x)-1=0 ?

The general solution for sin^2(x)-cos^2(x)-1=0 is x= pi/2+pin

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sin^2(x)-cos^2(x)-1=0

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

x = \frac{π}{2} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

= - 1 - cos (2 x)

- 1 - cos (2 x) = 0

Move

1

to the right side

- 1 - cos (2 x) = 0

Add

1

to both sides

- 1 - cos (2 x) + 1 = 0 + 1

Simplify

- cos (2 x) = 1

- cos (2 x) = 1

Divide both sides by

- 1

- cos (2 x) = 1

Divide both sides by

- 1

\frac{- cos ( 2 x )}{- 1} = \frac{1}{- 1}

Simplify

cos (2 x) = - 1

cos (2 x) = - 1

General solutions for

cos (2 x) = - 1

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

tan (x) - 1 = - \frac{1}{3} + \frac{1}{3} cot (x)

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

75 = 50 cosh (\frac{x}{50})

arcsin (\frac{1}{z}) = 0

3 sin^{2} (θ) - 4 = - 4 sin (θ)

What is the general solution for sin^2(x)-cos^2(x)-1=0 ?
The general solution for sin^2(x)-cos^2(x)-1=0 is x= pi/2+pin