2cos2x(sin2x) + sin2x = 0?

Question

I need help solving this problem, Solve the following trigonometric equation on the interval [ 0, 2pi ) : 2cos2x(sin2x) + sin2x = 0?

Answer 1

Hi Seth,

 

See the following graph to convince yourself that on the interval from [0, 2pi) this equation has 8 solutions.

 

https://www.desmos.com/calculator/4qqtxxk6wr

 

The question is how to you find those solutions algebraically?

 

Start by factoring out the sin(2x) from each term to reach:

 

sin2x[2cos2x + 1] = 0

 

By the zero product property either sin2x = 0 OR 2cos2x+1 = 0.

 

To solve sin2x = 0 we start by solving for 2x.

2x = arc sin 0 

The angles that have sin of 0 are 0 + kpi, where k is any integer.

That means x = kpi/2 , where k is any integer. Let k = 0, 1, 2, ... until you are outside the target interval.

So we have x = 0, pi/2, pi, 3pi/2.

 

Now let's work on 2cos2x +1 = 0.

We get cos2x = -1/2.

So 2x = arc cos -1/2.

The angles that have a cos of -1/2 are 2pi/3 + 2kpi and 4pi/3 + 2kpi, where k is any integer.

Once again let k = 0, 1, 2... until you are outside the target interval.

 

When 2x = 2pi/3 + 2kpi, x = pi/3 + kpi, or pi/3, 4pi/3 (k=0, 1)

When 2x = 4pi/3 + 2kpi, x= 2pi/3 + kpi, or 2pi/3, 5pi/3 (k=0,1)

 

There you go.

 

 

Answer 2

I don't think there is no general way to solve this equation; you need to proceed in steps.

 

First let us consider the situation where sin 2x = 0, i.e. at 0 and π/2.  There you have 2 roots.

 

Now look at all the values for which sin 2x ≠ 0.  At those values divide out sin 2x to get

2cos2x + 1 = 0 => cos 2x = -1.  cos x = -1 at only one place, namely where x = π  so cos 2x = -1 only when 2x = π, i.e. at x=π/2.   BUT we already saw that at x=π/2, the division is not allowed.  But it doesn't matter:

 

At π/2 the equation becomes 0 * 0 + 0 = 0.

 

I think you have only 2 roots as we have determined.