Question #74920

2 Answers
P dilip_k
Mar 15, 2017

#2sin^2x+sin2x=0#

#=>2sin^2x+2sinxcosx=0#

#=>2sinx(sinx+cosx)=0#

So #sinx=0=sin0=sinpi=sin2pi#

#=>x=0 or pi or 2pi#

#=>x=0^@ or 180^@ or 360^@#

Again

#=>sinx+cosx=0

#=>sinx=-cosx#

#=>sinx/cosx=-1

#=>tanx=-1#

#=>tanx=-tan(pi/4)=tan(pi-pi/4)=tan(2pi-pi/4)#

So

#x=(3pi)/4 or 135^@ and (7pi)/4 or 315^@#

Nghi N.
Mar 15, 2017

#0, (3pi)/4, pi, (7pi)/4, 3pi#

Explanation:

#2sin^2 x + 2sin x.cos x = 0#
sin x(sin x + cos x) = 0
a. sin x = 0 --> 3 solution arcs -->
x = 0, #x = pi#, #x = 2pi#
b. sin x + cos x = 0
From trig identity :
#sin x + cos x = sqrt2cos (x - pi/4)#, we get:
#cos (x - pi/4) = 0# --> 2 solutions -->
#x - pi/4 = pi/2# --> #x = pi/2 + pi/4 = (3pi)/4#
#x - pi/4 = (3pi/2)# --> #x = (3pi)/2 + pi/4 = (14pi)/8 = (7pi)/4#
Answers for #(0, 2pi)#:
#0, (3pi)/4, pi, (7pi)/4, 2pi#

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