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