Question 328059
Use the trig identitites along with your equation.
{{{sin(x)-cos(x)=1}}}
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{{{sin(x)=cos(x)+1}}}
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{{{sin^2(x)=(cos(x)+1)^2}}}
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{{{1-cos^2(x)=cos^2(x)+2cos(x)+1}}}
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{{{2cos^2(x)+2cos(x)=0}}}
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{{{cos(x)(cos(x)+1)=0}}}
Two solutions from the zero product property:
{{{cos(x)=0}}}
{{{x=pi/2}}} and {{{x=(3pi)/2}}}
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{{{cos(x)+1=0}}}
{{{cos(x)=-1}}}
{{{x=pi}}}
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Verify the solutions.
{{{sin(x)-cos(x)=1}}}
{{{sin(pi)-cos(pi)=1}}}
{{{0-(-1)=1}}}
{{{1=1}}}
{{{x=pi}}} is a valid solution.
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{{{sin(x)-cos(x)=1}}}
{{{sin(pi/2)-cos(pi/2)=1}}}
{{{1-0=1}}}
{{{1=1}}}
{{{x=pi/2}}} is a valid solution.
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{{{sin(x)-cos(x)=1}}}
{{{sin((3pi)/2)-cos((3pi)/2)=1}}}
{{{-1-0=1}}}
{{{-1=1}}}
{{{x=(3pi)/2}}} is not a valid solution.
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Two solution: {{{x=pi/2}}} and {{{x=pi}}}
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{{{drawing(300,300,-1,4,-2,2,circle(pi,0,.1),circle(1.57,0,.1),circle(pi/2,0,.1),circle(pi,1,.1),blue(line(pi,0,pi,1)),blue(line(pi/2,0,pi/2,1)),circle(pi/2,1,.1),circle(pi,1,.1),grid(1),graph(300,300,-1,4,-2,2,sin(x)-cos(x), 1))}}}