Solve x: Trig Equation (4/3)sinXcosX-sin22X=0

[Cuisine123]

Homework Statement

Solve for x. x is greater than or equal to 0 and x is less than or equal to 2pi.

(4/3)sinxcosx-sin²2x=0

If there is a better way to approach this question than the way I did it below, please post it.
Thanks.

Homework Equations

N/A

The Attempt at a Solution

(4/3)sinxcosx-sin²2x=0
2(2sinxcosx)-3sin²2x=0
2(sin2x)-3sin²2x=0
2sin2x-3sin²2x=0
sin2x(2-3sinx)=0 <------I know this is incorrect. How do I factor the previous line correctly?

To solve the first part:
sin2x=0
2x=sin^-1(0)=0,pi,2pi,3pi,4pi
therefore x=0, pi/2, pi, 3pi/2, 2pi

I know there are other solutions for x. Please help!

 

[ehild]

The Attempt at a Solution

(4/3)sinxcosx-sin²2x=0
2(2sinxcosx)-3sin²2x=0
2(sin2x)-3sin²2x=0
2sin2x-3sin²2x=0
sin2x(2-3sinx)=0 <------I know this is incorrect. How do I factor the previous line correctly?

You just left out the factor "2" in front of x : the second factor is 2-3sin(2x).

As you have found out, the first set of solutions is x= k*pi/2, with k integer.
You get the other set by equating the second factor to zero: 2-3sin(2x)=0

ehild

 

[Bohrok]

Looks like you can solve it using the substitution [STRIKE]u = tan x[/STRIKE] u = tan(x/2)

 

[Cuisine123]

Looks like you can solve it using the substitution u = tan x

How do I do that? Can you please explain it more clearly? Thanks.

 

[Cuisine123]

You just left out the factor "2" in front of x : the second factor is 2-3sin(2x).

As you have found out, the first set of solutions is x= k*pi/2, with k integer.
You get the other set by equating the second factor to zero: 2-3sin(2x)=0

ehild

So if I solve for x in 2-3sin(2x)=0, I'll get the following in rad: 0.365, 1.205, 3.505, 4.345?
Someone..please correct me if I am wrong.

 

[ehild]

Your numbers are correct but you have infinite number of solutions, it is not enough to write down some of them.

2x=0.730+k*2pi , or
2x=pi-0.730+2k*pi=2.41+2k*pi, with k an integer (positive or negative or zero).

So your solutions are in general form:

x=0.365 + k*pi,

x=1.206 + k*pi,

and x=k*pi/2 ,

with k = integer.

k can be negative, so x can be also -2.776 or -1.936 ...

ehild

 

[Bohrok]

I had a mistake in my last message which is corrected now.

Using the substitution u = tan(x/2) gives us sin x = 2u/(u² + 1) and cos x = (1 - u²)/(u² + 1). Substitute those into your equation and you can rewrite it into a quadratic equation in terms of u. See if you can solve for x after you substitute tan(x/2).

 

[vela]

Your numbers are correct but you have infinite number of solutions, it is not enough to write down some of them.

The problem statement restricts x to the interval [0,2\pi].

 

[ehild]

The problem statement restricts x to the interval [0,2\pi].

Ok, your solution is correct then.

ehild