How do you solve 5sin^2x-8sinx=35sin2x−8sinx=3 in the interval [0,360]?

1 Answer
Nghi N
May 1, 2018

90^@; 216^@87; 323^@1390∘;216∘87;323∘13

Explanation:

f(x) = 5sin^2 x - 8sin x - 3 = 0f(x)=5sin2x−8sinx−3=0
Solve the above quadratic equation for sin x.
Since a + b + c = 0, the shortcut gives:
sin x = 1 and sin x = c/a = - 3/5sinx=ca=−35
a. sin x = 1 --> x = pi/2 = 90^@x=π2=90∘
b. sin = -3/5sin=−35
Calculator and unit circle give 2 solutions:
x = - 36^@87x=−36∘87, or x = 360 - 36.87 = 323^@13x=360−36.87=323∘13 (co-terminal), and:
x = 180 - (- 36.87) = 216^@87x=180−(−36.87)=216∘87

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