What is the general solution for 3cos^2(θ)-3=sin^2(θ) ?

The general solution for 3cos^2(θ)-3=sin^2(θ) is θ=2pin,θ=pi+2pin

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3cos^2(θ)-3=sin^2(θ)

3 cos^{2} (θ) - 3 = sin^{2} (θ)

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

3 cos^{2} (θ) - 3 = sin^{2} (θ)

Subtract

sin^{2} (θ)

from both sides

3 cos^{2} (θ) - 3 - sin^{2} (θ) = 0

Rewrite using trig identities

- 3 - sin^{2} (θ) + 3 cos^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

- 3 - sin^{2} (θ) + 3 (1 - sin^{2} (θ)) : - 4 sin^{2} (θ)

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

Expand

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

3 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (θ)

= 3 \cdot 1 - 3 sin^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (θ)

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

Simplify

- 3 - sin^{2} (θ) + 3 - 3 sin^{2} (θ) : - 4 sin^{2} (θ)

- 3 - sin^{2} (θ) + 3 - 3 sin^{2} (θ)

Group like terms

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

Add similar elements:

- sin^{2} (θ) - 3 sin^{2} (θ) = - 4 sin^{2} (θ)

= - 4 sin^{2} (θ) - 3 + 3

- 3 + 3 = 0

= - 4 sin^{2} (θ)

= - 4 sin^{2} (θ)

= - 4 sin^{2} (θ)

- 4 sin^{2} (θ) = 0

Divide both sides by

- 4

- 4 sin^{2} (θ) = 0

Divide both sides by

- 4

- 4 sin^{2} (θ) = 0

Divide both sides by

- 4

\frac{- 4 sin ^{2} ( θ )}{- 4} = \frac{0}{- 4}

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

y, ln (sin^{2} (3 x)) = e^{x} + arccot (y)

tan (x) = a

3 tan^{2} (x) = 2 sec^{2} (x) + 1

cos (A) = \frac{8}{15}

- 6 cos (θ) = 2 - 2 cos (θ)

What is the general solution for 3cos^2(θ)-3=sin^2(θ) ?
The general solution for 3cos^2(θ)-3=sin^2(θ) is θ=2pin,θ=pi+2pin