SOLUTION: Find all the solutions [in degrees] to 24 sin2𝜃 − 5 sin 𝜃 = 1 Round your answers to one decimal place.

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Question 759571: Find all the solutions [in degrees] to 24 sin2𝜃 − 5 sin 𝜃 = 1 Round your answers to one decimal place.
Answer by lwsshak3(11628) About Me  (Show Source):
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Find all the solutions [in degrees] to 24 sin2𝜃 − 5 sin 𝜃 = 1 Round your answers to one decimal place.
24 sin^2𝜃 − 5 sin 𝜃 - 1=0
solve for sin 𝜃 using quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=24, b=-5, c=-1
sin 𝜃≈-0.125
𝜃≈187.2º+360ºn, 352.8º+360ºn, n=integer (in quadrants III and IV where sin<0)
or
sin𝜃≈0.333
𝜃=19.5º+360ºn,160.5º+360ºn, n=integer (in quadrants I and II where sin>0)