(A) Given that 2sin^2(θ) + sin(θ) - 1 = 0, find the two values for sin θ

(B) Given that 0° ≤ θ ≤ 360° and that one solution for θ is 30°, find the other two possible values for θ.

Guest Dec 10, 2019

edited by
Guest
Dec 10, 2019

#1**0 **

A: Let x = sin (theta)

then 2x^2 +x -1 = 0 then use quadratic formula to find x (which is sin(theta))

B: Use your answer from A to determine theta possibilities

ElectricPavlov Dec 10, 2019

#3**0 **

If you do it correctly, you should find x = 1/2 or -1 (which is the value of sin(theta) )

ElectricPavlov
Dec 10, 2019