# How do I solve the equation "2cos(x) = sin(2x), for 0 ≤ x ≤ 3π"?

The key to solving this equation is realizing that sin(2x) can be written in terms of sin(x) and cos(x) using a **double-angle formula**. (With trigonometric problems similar to this one, you should always check if any trigonometric identites like the double-angle formulae can be used, as these can often help you.)

Using your IB formula booklet, you will see that the double-angle formula for sine is:

sin(2x) = 2sin(x)cos(x)

Therefore, we can rewrite the given equation from:

2cos(x) = sin(2x)

to:

2cos(x) = 2sin(x)cos(x)

Next, we notice that both sides of the equation are multiplied by 2, so we can divide both sides by 2. This yields the equation:

cos(x) = sin(x)cos(x)

We can now divide both sides of the equation by cos(x), which leaves us with:

sin(x) = 1

Finally, we must think about the angles at which sin(x) is equal to 1. You should realize, perhaps by imagining the unit circle, that sine is equal to 1 whenever x = π/2 + n2π, where n is any integer.

However, this is not the final answer, as the problem gave us a restricted domain for x. x must be in between 0 and 3π. Therefore, the only possible values for x are π/2 and 5π/2.

So, the answer is:

x = π/2 and 5π/2