Question 759571
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=24, b=-5, c=-1
sin 𝜃≈-0.125
&#120579;&#8776;187.2º+360ºn, 352.8º+360ºn, n=integer (in quadrants III and IV where sin<0)
or
sin&#120579;&#8776;0.333
&#120579;=19.5º+360ºn,160.5º+360ºn, n=integer  (in quadrants I and II where sin>0)