find all solutions of the equation 2sinxcos2x-cos2x=0 over the interval 0<=pi

Answer 1

To find all solutions of the equation 2sinx*cos(2x) - cos(2x) = 0 over the interval 0 <= x <= pi, we can factor out cos(2x) from the equation:

cos(2x)(2sinx - 1) = 0

This means that either cos(2x) = 0 or 2sinx - 1 = 0.

1. To find the solutions for cos(2x) = 0, we know that cos(2x) = 0 when 2x = pi/2 + n*pi where n is an integer. Therefore, x = (pi/2 + n*pi)/2 for n = 0, 1. We need to check if these solutions fall within the interval [0, pi].

For n = 0, x = pi/4. This falls within the interval [0, pi], so x = pi/4 is a solution.

For n = 1, x = (3pi/2)/2 = 3pi/4. This also falls within the interval [0, pi], so x = 3pi/4 is a solution.

2. To find the solutions for 2sinx - 1 = 0, we have sinx = 1/2. The solutions of sinx = 1/2 are x = pi/6 and x = 5pi/6. We need to check if these solutions also fall within the interval [0, pi].

Both solutions x = pi/6 and x = 5pi/6 are within the interval [0, pi].

Therefore, the solutions of the equation 2sinx*cos(2x) - cos(2x) = 0 over the interval 0 <= x <= pi are x = pi/4, 3pi/4, pi/6, and 5pi/6.