Posted by Joe on Saturday, July 18, 2009 at 11:10pm.
how can i find all the exact solutions to the equation : 2cos^2x + 3sinx = 3
the solutions have to be between [0,2pi)

trig  Quidditch, Saturday, July 18, 2009 at 11:41pm
assuming that 2cos^2x means 2(cos(x))^2
substitute using
(cos(x))^2 = 1  (sin(x))^2
2(1  (sin(x))^2) + 3sin(x) = 3
2  2(sin(x))^2 + 3sin(x) = 3
for clarity, let u = sin(x)
2  2u^2 + 3u = 3
rearranging...
2u^2  3 u + 1 = 0
Solve this quadratic to find the values of sin x that are solutions. Finally, determine the angles in the range that have that value of sin(x).
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