What is the general solution for cos^2(θ)=sin^2(θ)+1,0<= θ<= 360 ?

The general solution for cos^2(θ)=sin^2(θ)+1,0<= θ<= 360 is θ=0,θ=180,θ=360

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cos^2(θ)=sin^2(θ)+1,0<= θ<= 360

cos^{2} (θ) = sin^{2} (θ) + 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

Radians

θ = 0, θ = π, θ = 2 π

Solution steps

cos^{2} (θ) = sin^{2} (θ) + 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

Subtract

sin^{2} (θ) + 1

from both sides

cos^{2} (θ) - sin^{2} (θ) - 1 = 0

Rewrite using trig identities

cos^{2} (θ) - sin^{2} (θ) - 1

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

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 + 36 0^{\circ} n

2 θ = 0 + 36 0^{\circ} n

Solve

2 θ = 0 + 36 0^{\circ} n : θ = 18 0^{\circ} n

2 θ = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

2 θ = 36 0^{\circ} n

Divide both sides by

2

2 θ = 36 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{36 0 ^{\circ} n}{2}

Simplify

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

θ = 18 0^{\circ} n

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 0, θ = 18 0^{\circ}, θ = 36 0^{\circ}

tan (2 x) (1 - tan^{2} (x)) = \frac{2}{3}

tan (x) = 25762

sin (8 x) = sin (6 x)

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

0 = sec (\frac{π x}{4})

What is the general solution for cos^2(θ)=sin^2(θ)+1,0<= θ<= 360 ?
The general solution for cos^2(θ)=sin^2(θ)+1,0<= θ<= 360 is θ=0,θ=180,θ=360