Question 1077821
a) {{{sin(x+pi/3)+sin(x-pi/3)=1}}}
Using the trigonometric identities
for sine of sum an difference of two angles,
the equation can be re-written as
{{{sin(x)cos(pi/3)+cos(x)sin(pi/3)}}}{{{"+"}}}{{{sin(x)cos(pi/3)-cos(x)sin(pi/3)=1}}}
Taking out {{{sin(x)}}} and {{{sin(x)}}} as common factors
the equation can be re-written as
{{{sin(x)(cos(pi/3)+cos(pi/3))}}}{{{"+"}}}{{{cos(x)(sin(pi/3)-sin(pi/3))=1}}}
{{{sin(x)(cos(pi/3)+cos(pi/3))}}}{{{"+"}}}{{{cos(x)(0)=1}}}
{{{sin(x)(cos(pi/3)+cos(pi/3))=1}}}
We know that {{{cos(pi/3)=1/2}}} , so we re-write the equation as
{{{sin(x)(1/2+1/2)=1}}} and {{{sin(x)=1}}} .
In the interval [0,2pi), that happens only for
{{{x=pi/2}}} .
{{{drawing(600,200,-0.5,6.5,-1.2,1.3,graph(600,200,-0.5,6.5,-1.2,1.3,sin(x+pi/3)+sin(x-pi/3),1),circle(pi/2,1,0.02),locate(1.5,0.97,pi/2))}}}
 
b) {{{tan(x+pi)+2sin(x+pi)=0}}}
Based on trigonometric identities, the equation can be re-written as
{{{tan(x)-2sin(x)=0}}} and {{{sin(x)/cos(x)-2sin(x)=0}}} .
Then, with some algebra, it can be re-written as
{{{sin(x)(1/cos(x)-2)=0}}} and {{{sin(x)(1-2cos(x))/cos(x)=0}}}
The numerator is zero when
{{{sin(x)=0}}} ---> {{{highlight(x=0)}}} or {{{highlight(x=pi)}}} .
The numerator is also zero when
{{{1-2cos(x)=0}}} ---> {{{cos(x)=1/2}}} ---> {{{highlight(x=pi/3)}}} or {{{highlight(x=5pi/3)}}} .
For none of those values of x, is the {{{cos(x)}}} zero,
so they are all valid solutions.
{{{drawing(600,200,-0.5,6.5,-1.2,1.2,graph(600,200,-0.5,6.5,-1.2,1.2,tan(x+pi)+2sin(x+pi)),circle(0,0,0.03),locate(0.03,0.2,0),
circle(pi,0,0.03),locate(3.2,0.3,pi),
circle(pi/3,0,0.03),locate(1.05,0,pi/3),
circle(5pi/3,0,0.03),locate(5.3,0,5pi/3)
)}}}
 
c) {{{cos(x-pi/2)+sin(2x)=0}}}
(Or did you mean {{{cos(x-pi/2)+sin^2(x)=0}}} instead?)
Using trigonometric identities,
the equation can be re-written as
{{{cos(-(x-pi/2))+sin(2x)=0}}} <--> {{{cos(pi/2-x)+sin(2x)=0}}} and {{{sin(x)+sin(2x)=0}}} .
If the second term was really {{{}}} ,
using the trig identity for double angles,
the equation can be re-written as
{{{sin(x)+2sin(x)cos(x)=0}}} <---> {{{sin(x)(1+2cos(x))=0}}}
The expression {{{sin(x)(1+2cos(x))}}} is zero when
{{{sin(x)=0}}} ---> {{{highlight(x=0)}}} or {{{highlight(x=pi)}}} .
The expression {{{sin(x)(1+2cos(x))}}} is also zero when
{{{1+2cos(x)=0}}} <---> {{{cos(x)=-1/2}}} .
In the interval [0,2pi), that happens for
{{{highlight(x=2pi/3)}}} or {{{highlight(x=4pi/3)}}} .
{{{drawing(600,260,-0.5,6.5,-1.8,1.8,graph(600,260,-0.5,6.5,-1.8,1.8,cos(x-pi/2)+sin(2x)),circle(0,0,0.03),locate(0.05,0.2,0),
circle(pi,0,0.03),locate(3.05,0.25,pi),
circle(2pi/3,0,0.03),locate(2.1,0.5,2pi/3),
circle(4pi/3,0,0.03),locate(4.2,0.5,4pi/3)
)}}}
 
NOTE: For {{{cos(x-pi/2)+sin^2(x)=0}}} ,
{{{sin(x)+sin^2(x)=0}}} <--> {{{sin(x)(1+sin(x))=0}}} ,
in the interval [0,2pi) has solutions when
{{{sin(x)=0}}} --> {{{x=0}}} or {{{x=pi}}} ,
and when
{{{1+sin(x)=0}}} --> {{{sin(x)=-1}}} --> {{{x=3pi/2}}} .