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

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

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cos^2(θ)-1=sin^2(θ)

cos^{2} (θ) - 1 = sin^{2} (θ)

θ = πn

Degrees

θ = 0^{\circ} + 18 0^{\circ} n

Solution steps

cos^{2} (θ) - 1 = sin^{2} (θ)

Subtract

sin^{2} (θ)

from both sides

cos^{2} (θ) - 1 - sin^{2} (θ) = 0

Rewrite using trig identities

cos^{2} (θ) - 1 - sin^{2} (θ)

Use the Double Angle identity:

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

= - 1 + cos (2 θ)

- 1 + cos (2 θ) = 0

Move

1

to the right side

- 1 + cos (2 θ) = 0

Add

1

to both sides

- 1 + cos (2 θ) + 1 = 0 + 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

- 3 cos^{2} (u) sin (u) - 2 sin (2 u) = 0

cot (θ) = - \frac{7}{24}

2 sin (x) + 2 cos (x) = 6

0 = - 0.65 sin (13 x)

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

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