Posted by Calvin on Thursday, May 17, 2012 at 4:47pm.
solve 2cos^2xsinx=1, 0 < or equal to x < 2pie

trig  Bosnian, Thursday, May 17, 2012 at 5:33pm
cos ^ 2 ( x ) = 1  sin ^ 2 ( x )
2 cos ^ 2 ( x )  sin ( x ) = 1
2 * [ 1  sin ^ 2 ( x ) ]  sin ( x ) = 1
2  2 sin ^ 2 ( x )  sin ( x ) = 1 Subtract 1 to both sides
2  2 sin ^ 2 ( x )  sin ( x )  1 = 1  1
 2 sin ^ 2 ( x )  sin ( x ) + 1 = 0
Substitute :
sin ( x ) = u
 2 u ^ 2  u + 1 = 0
The exact solution are :
u = 1 / 2
u =  1
cos ( x ) = 1 / 2
cos ( x ) =  1
x = 30 ° = pi / 6
x = 270 ° = 9 pi / 6 rad
P.S
If you don't know how to solve equation :
 2 u ^ 2  u + 1 = 0
In google type:
quadratic equation online
When you see list of results click on:
Free Online Quadratic Equation Solver:Solve by Quadratic Formula
When page be open in rectangle type:
 2 u ^ 2  u + 1 = 0
and click option: solve it!
You wil see solution stepbystep

trig  Bosnian, Thursday, May 17, 2012 at 5:34pm
Corection:
sin ( x ) = 1 / 2
sin ( x ) =  1

trig  ky, Thursday, May 17, 2012 at 5:43pm
would the 3 answers be: pi/6, 5pi/6, 3pi/2

trig  Bosnian, Thursday, May 17, 2012 at 6:00pm
Yes
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