How do I solve given #0<x<2pi# #2sinxcosx=sqrt2cosx#?

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Answer 1 · 2018-04-24T00:38:10

# x in { pi/4, pi/2, {3pi}/4, {3pi}/2 } #

#2 sin x cos x = sqrt{2} cos x#

#2 sin x cos x - sqrt{2} cos x = 0#

# cos x(2 sin x - sqrt{2}) = 0#

#cos x = 0# or #sin x = \sqrt{2}/2#

#x = 90^circ + 180^circ k quad# for integer #k# or

#x = 45^circ + 360^circ k# or

# x = 135^circ + 360^circ k#

In radians from #0# to #2pi# that's

# x = { pi/4, pi/2, {3pi}/4, {3pi}/2 } #

Check:

#2 sin x cos x = sin(2x)#

# x=pi/4, quad sin(2x)=sin(pi/2)=1 #, #sqrt 2 cos (pi/4) = 1 quad sqrt#

# x=pi/2, quad sin(2x)=0#, #sqrt 2 cos (pi/2) = 0 quad sqrt#

I'll leave the other to you to check.