Friday

April 18, 2014

April 18, 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)

**Related Questions**

Differential Equations - Consider the differential equation: dy/dt=y/t^2 a) Show...

Differential Equations - Consider the differential equation: dy/dt=y/t^2 a) Show...

Differential Equations - Consider the differential equation: dy/dt=y/t^2 a) Show...

Precalculs - I have no idea how to do these type of problems. -------Problem...

solution of an equation - can someone please give me an example of a solution of...

trig - find each value by referring to the graphs of the trig functions. 1. sec ...

Trig - Find all solutions of the equation in the interval [0,2pi). sqrt(3)...

advanced functions - Solve 2tan^2x + 1 = 0 Interval : [0, 2 pi]

Advanced Math - Find the values of x for which the equation sin x = -1 is true 2...

Trig - Solve the equation on the interval 0 less than or equal 0 < 2pi. 1. ...