Question

2sinxcosx=sinx in the interval of [0,2pi]

Answer 1

Use the double angle function sin2x = 2sinxcosx.

2sinxcosx + sinx = 0

For the first part of the problem, consider the solutions on the interval where sinx = 0

x = 0, pi

Now consider the solutions where sinx does NOT equal zero.

2sinxcosx + sinx = 0

2sinxcosx = - sinx

2cosx = - 1

cosx = - 1/2

x = 2pi/3, 4pi/3