What is the general solution for 2sin^2(θ)-5sin(θ)+3=0 ?

The general solution for 2sin^2(θ)-5sin(θ)+3=0 is θ= pi/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(θ)-5sin(θ)+3=0

2 sin^{2} (θ) - 5 sin (θ) + 3 = 0

θ = \frac{π}{2} + 2 πn

Solution steps

2 sin^{2} (θ) - 5 sin (θ) + 3 = 0

Solve by substitution

sin (θ) = \frac{3}{2}, sin (θ) = 1

sin (θ) = \frac{3}{2} :

No Solution

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn

2 sin (x) csc (x) - csc (x) = 4 sin (x) - 2

2 cos^{2} (x) + 5 cos (x) + 2 = 0

sin (θ) = - 0.66

4 sec^{2} (θ) tan (θ) = 8 tan (θ)

5 sec (x) csc (x) = 10 csc (x)

What is the general solution for 2sin^2(θ)-5sin(θ)+3=0 ?
The general solution for 2sin^2(θ)-5sin(θ)+3=0 is θ= pi/2+2pin