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

The general solution for sin^2(θ)-cos^2(θ)=-1 is θ=pin

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sin^2(θ)-cos^2(θ)=-1

sin^{2} (θ) - cos^{2} (θ) = - 1

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

sin^{2} (θ) - cos^{2} (θ) = - 1

Rewrite using trig identities

sin^{2} (θ) - cos^{2} (θ)

Use the Double Angle identity:

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

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

= - cos (2 θ)

- cos (2 θ) = - 1

Divide both sides by

- 1

- cos (2 θ) = - 1

Divide both sides by

- 1

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

Simplify

cos (2 θ) = 1

cos (2 θ) = 1

General solutions for

cos (2 θ) = 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 θ = 0 + 2 πn

2 θ = 0 + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

Divide both sides by

2

2 θ = 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2}

Simplify

θ = πn

θ = πn

θ = πn

cos^{2} (x) - 4 cos (x) = 0

x, sin (x) + cot (x) cos (x) = \frac{1}{f}

\frac{1}{sin ( x )} = 2

tan (x) csc (x) - 2 tan (x) = 0

1 - 2 sin (θ) = 0

What is the general solution for sin^2(θ)-cos^2(θ)=-1 ?
The general solution for sin^2(θ)-cos^2(θ)=-1 is θ=pin