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

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

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cos(2θ)-sin^2(θ)=1

cos (2 θ) - sin^{2} (θ) = 1

θ = 2 πn, θ = π + 2 πn

Degrees

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

Solution steps

cos (2 θ) - sin^{2} (θ) = 1

Subtract

1

from both sides

cos (2 θ) - sin^{2} (θ) - 1 = 0

Rewrite using trig identities

- 1 + cos (2 θ) - sin^{2} (θ)

Use the Double Angle identity:

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

= - 1 + 1 - 2 sin^{2} (θ) - sin^{2} (θ)

Simplify

- 1 + 1 - 2 sin^{2} (θ) - sin^{2} (θ) : - 3 sin^{2} (θ)

- 1 + 1 - 2 sin^{2} (θ) - sin^{2} (θ)

Add similar elements:

- 2 sin^{2} (θ) - sin^{2} (θ) = - 3 sin^{2} (θ)

= - 1 + 1 - 3 sin^{2} (θ)

- 1 + 1 = 0

= - 3 sin^{2} (θ)

= - 3 sin^{2} (θ)

- 3 sin^{2} (θ) = 0

Divide both sides by

- 3

- 3 sin^{2} (θ) = 0

Divide both sides by

- 3

- 3 sin^{2} (θ) = 0

Divide both sides by

- 3

\frac{- 3 sin ^{2} ( θ )}{- 3} = \frac{0}{- 3}

Simplify

sin^{2} (θ) = 0

sin^{2} (θ) = 0

Apply rule

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

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (x) = \frac{3}{4}

2 sin (2 x) = 3 sin (x)

4 sin (\frac{t}{2}) - 1 = 1, - \frac{π}{2} \leq t \leq \frac{π}{2}

- 6 cos (θ) - 10 = - 7

sin^{2} (x) = cos^{2} (x) + 1

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