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(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.

 Dec 10, 2019
edited by Guest  Dec 10, 2019
 #1
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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

 Dec 10, 2019
edited by ElectricPavlov  Dec 10, 2019
 #3
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If you do it correctly, you should find x = 1/2   or  -1     (which is the value of sin(theta) )

ElectricPavlov  Dec 10, 2019
 #2
avatar+11646 
+1

(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.

 

laugh

 

 Dec 10, 2019
edited by Omi67  Dec 10, 2019
edited by Omi67  Dec 10, 2019

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