What is the general solution for 6+6sin(θ)=4cos^2(θ) ?

The general solution for 6+6sin(θ)=4cos^2(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6+6sin(θ)=4cos^2(θ)

6 + 6 sin (θ) = 4 cos^{2} (θ)

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

6 + 6 sin (θ) = 4 cos^{2} (θ)

Subtract

4 cos^{2} (θ)

from both sides

6 + 6 sin (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

2 + 4 sin^{2} (θ) + 6 sin (θ) = 0

Solve by substitution

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

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

5 cos (x) + 5 sin (x) tan (x) = 10

sin (x) = 0.33

2 cos (2 x) = - 1

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

5 sin (2 θ) - 7 sin (θ) = 0

What is the general solution for 6+6sin(θ)=4cos^2(θ) ?
The general solution for 6+6sin(θ)=4cos^2(θ) is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=(3pi)/2+2pin