How do you solve #sin2x-cosx=0#?

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Answer 1 · 2016-11-25T18:52:04

#x = pi/2, (3pi)/2, pi/6, (5pi)/6#

Before we solve, we need to note an identity:

#sin2x = 2sinxcosx#

Using the identity from above, rewrite the equation.

#2sinxcosx - cosx=0#

Now factor out a #cosx#.

#cosx(2sinx - 1) = 0#

Now take each factor and set it equal to zero.

#cosx = 0#

#x = pi/2, (3pi)/2#

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#2sinx-1 = 0#

#2sinx = 1#

#sinx = 1/2#

#x = pi/6, (5pi)/6#