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November 28, 2014

November 28, 2014

Posted by **Anonymous** on Monday, November 16, 2009 at 5:24pm.

4sinxcos2x + 4cosxsin2x - 1 = 0

in the Interval [0, 2 pi]

Either show that there is no solution to the equation in this domain, or determine the smallest possible solution.

- Advanced Functions -
**Reiny**, Monday, November 16, 2009 at 6:33pm4sinxcos2x + 4cosxsin2x - 1 = 0

4(sinxcos2x + cosxsin2x) = 1

Did you recognize that the expression in my bracket matches with

sin(A+B) = sinAcosB +cosAsinB ?

so we have

4(sin(x+2x)) = 1

sin 3x = 1/4

3x = .25268 radians

x = .084227 radians

(you wanted the smallest possible solution)

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