Solve for x by factoring

Question

Solve for x by factoring where 0≤x≤2∏

 

√3 sinx tanx=3 sinx

Answer 1

You are told to factor this equation, so approach it like anything that you would in Algebra. First, put everything to one side so that you have a bunch of stuff equal to zero.

 

√(3) sin(x) tan (x) - 3 sin(x) = 0

 

Now factor out anything that you can

 

sin(x) [√(3) tan(x) - 3] = 0

 

Just like when you used to factor polynomials, this is equal to 0 when either of the factored parts are equal to zero. Specifically, this equation is equal to zero when

 

sin(x) = 0

and/or

√(3) tan(x) - 3 = 0

 

For the first solution, sin(x) = 0 when x = 0, π, and 2π (because your domain was restricted to 0 ≤ x ≤ 2π)

 

For the second solution, you need to do a little bit more work. You need to solve for x, here, so we do the normal algebra operations until we get to tan(x) = ....

 

tan(x) = 3/√(3)

 

alternatively,

 

tan(x) = √(3)    you get this by rationalizing the denominator (aka multiplying by √(3)/√(3) )

 

Then we find x by taking the inverse tangent of both sides. This is a good idea because taking the inverse tangent of a tangent function gives us the argument of the original tangent. That may have been a bit wordy, but think of how you would "undo" the square root function - you would square it, right?

 

tan^-1 [tan(x)] = tan^-1 [ √(3) ]

x = tan^-1 [ √(3) ]

 

You can find the numerical answer by plugging this into your calculator. Just make sure that it is in your domain of 0 ≤ x ≤ 2π

 

Major note: The inverse trig function, written as ^-1 is NOT the same thing as the reciprocal (1/whatever). These are fundamentally different things.

Answer 2

The way the equation is currently stated, you have

 

√3 sin(x) tan(x) = 3 sin(x)

 

If sin(x) ≠ 0,

 

√3 tan(x) = 3

 

tan(x) = 3/√3 = √3

 

x = tan^-1(√3)

 

If sin(x) = 0, then x = 2πn, n = 0,±1,±2,±3, ...