solve for sinx=cos2x (between 0 and 2pi)

is that cos(2x) or cos^2 (x) ?

cos(2x)

To solve the equation sin(x) = cos(2x), you need to find the values of x that satisfy this equation between 0 and 2π.

Let's break this down step by step:

1. Start by rewriting cos(2x) in terms of sine using the double-angle identity:
cos(2x) = 1 - 2sin^2(x)

Now, we have the equation sin(x) = 1 - 2sin^2(x)

2. Rearrange the equation:
2sin^2(x) + sin(x) - 1 = 0

We now have a quadratic equation in terms of sin(x). Let's solve this using the quadratic formula.

3. Apply the quadratic formula:
sin(x) = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, a = 2, b = 1, and c = -1. Substituting these values, we have:
sin(x) = (-1 ± sqrt(1^2 - 4(2)(-1))) / (2(2))

4. Simplify the equation:
sin(x) = (-1 ± sqrt(1 + 8)) / 4
sin(x) = (-1 ± sqrt(9)) / 4
sin(x) = (-1 ± 3) / 4

5. Solve for the two possible solutions:
a) For sin(x) = (-1 + 3) / 4 = 2 / 4 = 1/2:
This implies x = arcsin(1/2) = π/6 or x = π - arcsin(1/2) = 5π/6

b) For sin(x) = (-1 - 3) / 4 = -4 / 4 = -1:
Since sin(x) cannot be greater than 1 or less than -1, there are no solutions in this case.

Hence, the solutions to the equation sin(x) = cos(2x) between 0 and 2π are:
x = π/6 and x = 5π/6.