Solve the question: sin(2x)=cos(x), for values of x between 0 and 2 Pi

You'll find it very helpful to remember that sin(2x)=2sin(x)cos(x) here, since this means that you're trying to find all the solutions of 2sin(x)cos(x)=cos(x). These will be equal under two possible scenarios: firstly if cos(x)=0 (since both sides will be zero), and secondly if 2sin(x)=1, or sin(x)=0.5 (i.e. after you've cancelled out the cos(x) terms from both sides). Solve both of those equalities for all possible values between 0 and 2pi, and you've done it. I think you'll find there are four of them - two from each scenario.

To solve the equation sin(2x) = cos(x), we need to find the values of x that satisfy this equation, in the given range of 0 to 2π.

First, we'll convert sin(2x) and cos(x) into their equivalent expressions using trigonometric identities.

Recall the double-angle identity for sine:
sin(2θ) = 2sin(θ)cos(θ)

Using this identity, we can rewrite sin(2x) as:
2sin(x)cos(x) = cos(x)

Now, let's simplify the equation:
2sin(x)cos(x) - cos(x) = 0

Factoring out cos(x), we have:
cos(x)(2sin(x) - 1) = 0

Now we have two possibilities:
1) cos(x) = 0
2) 2sin(x) - 1 = 0

Let's solve these individually:

1) cos(x) = 0
In the given range of 0 to 2π, cos(x) equals zero at x = π/2 and x = 3π/2.

2) 2sin(x) - 1 = 0
Solving this equation for sin(x):
2sin(x) = 1
sin(x) = 1/2

In the given range, sin(x) equals 1/2 at x = π/6 and x = 5π/6.

Therefore, the values of x that satisfy the equation sin(2x) = cos(x), in the range 0 to 2π, are x = π/2, 3π/2, π/6, and 5π/6.