What is the general solution for 7sin^2(θ)-15sin(θ)+8=0 ?

The general solution for 7sin^2(θ)-15sin(θ)+8=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 >

7sin^2(θ)-15sin(θ)+8=0

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

θ = \frac{π}{2} + 2 πn

Solution steps

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

Solve by substitution

sin (θ) = \frac{8}{7}, sin (θ) = 1

sin (θ) = \frac{8}{7} :

No Solution

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn

sin (5 t) = 0

cos (x) = 0, - π \leq x \leq π

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

sin (θ) = - \frac{4}{5}, \frac{7 π}{2} \leq θ < 4 π, cos (2 θ)

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

What is the general solution for 7sin^2(θ)-15sin(θ)+8=0 ?
The general solution for 7sin^2(θ)-15sin(θ)+8=0 is θ= pi/2+2pin