How do you solve #3sinx=3-3cosx# for #0<=x<=2pi#?

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Answer 1 · 2016-11-12T21:31:59

#{0, pi/2}#

Alternative Approach:

#3sinx + 3cosx = 3#

#3(sinx + cosx) = 3#

#sinx + cosx = 1#

#(sinx+ cosx)^2 = 1^2#

#sin^2x+ 2sinxcosx + cos^2x = 1#

#1 + 2sinxcosx = 1#

#2sinxcosx= 0#

#(sinx)(cosx) = 0#

#x = 0, pi/2, pi, (3pi)/2, 2pi#

Checking in the original equation, you will find that only #x = pi/2# and #x= 0# are proper solutions.

Hopefully this helps!