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July 30, 2014

July 30, 2014

Posted by **Joe** on Saturday, July 18, 2009 at 11:10pm.

the solutions have to be between [0,2pi)

- trig -
**Quidditch**, Saturday, July 18, 2009 at 11:41pmassuming 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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