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

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

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sin^2(θ)+2cos^2(θ)=2

sin^{2} (θ) + 2 cos^{2} (θ) = 2

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

sin^{2} (θ) + 2 cos^{2} (θ) = 2

Subtract

2

from both sides

sin^{2} (θ) + 2 cos^{2} (θ) - 2 = 0

Rewrite using trig identities

- 2 + sin^{2} (θ) + 2 cos^{2} (θ)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

2 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (θ)

= 2 \cdot 1 - 2 sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (θ)

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

Simplify

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

- 2 + sin^{2} (θ) + 2 - 2 sin^{2} (θ)

Group like terms

= sin^{2} (θ) - 2 sin^{2} (θ) - 2 + 2

Add similar elements:

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

= - sin^{2} (θ) - 2 + 2

- 2 + 2 = 0

= - sin^{2} (θ)

= - sin^{2} (θ)

= - sin^{2} (θ)

- sin^{2} (θ) = 0

Divide both sides by

- 1

- sin^{2} (θ) = 0

Divide both sides by

- 1

- sin^{2} (θ) = 0

Divide both sides by

- 1

\frac{- sin ^{2} ( θ )}{- 1} = \frac{0}{- 1}

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

cos (θ) = - c

sin (t) = 1

(tan (x) - 1) (3 tan (x) - 1) = 0

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

2 sin (- 3 x - \frac{π}{4}) = 2, 0 \leq x \leq 2 π

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