Doug C. answered • 05/03/18

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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.