Friday

March 27, 2015

March 27, 2015

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)

**Answer this Question**

**Related Questions**

pre-calculus - solve the equation 1. In x + In (x+12) =2 x= ? or either there is...

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...

advanced functions - Determine approximate solutions for this equation in the ...

composite functions - f(x)=2x-4, g(x)=x squared + 5x What is F o G and Domain ...

Math - cos x = -0.6 , 0 <= theta <= 2pi Use inverse trigonometric ...

advanced functions - solve the first equation in each pair of equations or y and...

Math - hello I need help to do this exercise Consider the function f (x) = (16 x...

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