I've run into trouble applying this technique on this equation: Sin[x]+Cos[x]==0. I know that the answer is derivable by moving the cosine on one side of the equation, squaring both sides, and using the Pythagorean Identity Sin^2[x] + Cos^2[x] == 1.
However, applying the cofunction technique yields this:
Code: Select all
Sin[x]+Cos[x]==0
Sin[x]==-Cos[x]
Sin[x]==-Sin[Pi/2-x]
Sin[x]==Sin[x-Pi/2]
x==x-Pi/2+2Pi*k
Pi/2==2Pi*k