How do you solve #sin2x-cosx=0#?
1 Answer
Nov 25, 2016
Explanation:
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(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#