What is the general solution for 3sin^2(θ)+4cos^2(θ)=4 ?

The general solution for 3sin^2(θ)+4cos^2(θ)=4 is θ=2pin,θ=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3sin^2(θ)+4cos^2(θ)=4

3 sin^{2} (θ) + 4 cos^{2} (θ) = 4

θ = 2 πn, θ = π + 2 πn

Solution steps

3 sin^{2} (θ) + 4 cos^{2} (θ) = 4

Subtract

4

from both sides

3 sin^{2} (θ) + 4 cos^{2} (θ) - 4 = 0

Rewrite using trig identities

- sin^{2} (θ) = 0

Divide both sides by

- 1

sin (θ) = 0

General solutions for

sin (θ) = 0

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 2 πn, θ = π + 2 πn

tan^{2} (x) - 3 tan (x) - 2 = 0

sin (3 x) + 1 = cos (3 x)

6 sin^{2} (x) = 5

tan (θ) = \frac{46}{33}

arcsin (y) = \frac{1}{2}

What is the general solution for 3sin^2(θ)+4cos^2(θ)=4 ?
The general solution for 3sin^2(θ)+4cos^2(θ)=4 is θ=2pin,θ=pi+2pin