How do you solve #cos x = sin 2x#, within the interval [0,2pi)?

1 Answer
Jim G.
Apr 9, 2016

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

Explanation:

Using the #color(blue)" double angle formula " #

sin2A = 2sinAcosA

hence: cosx = 2sinxcosx

and 2sinxcosx - cosx = 0

take out the common factor cosx

cosx( 2sinx - 1 ) = 0

We now have : cosx = 0 or 2sinx - 1 = 0 → # sinx = 1/2 #

solve : cosx = 0 # rArr x = pi/2 , x = (3pi)/2 #

solve # sinx = 1/2 rArr x = pi/6 , x = (5pi)/6 #

In the interval # [ 0 , 2pi ) #

solutions are #color(red)(|bar(ul(color(white)(a/a)color(black)(pi/6 ,pi/2 , (5pi)/6 , (3pi)/2)color(white)(a/a)|)))#

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