Goniometrische vergelijkingen

(Exacte waardencirkel)
\(4x+\frac{1}{3}\pi =k⋅\pi \)

\(4x=-\frac{1}{3}\pi +k⋅\pi \)
\(x=-\frac{1}{12}\pi +k⋅\frac{1}{4}\pi \)

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{6}\pi ∨x=\frac{5}{12}\pi ∨x=\frac{2}{3}\pi ∨x=\frac{11}{12}\pi ∨x=1\frac{1}{6}\pi ∨x=1\frac{5}{12}\pi ∨x=1\frac{2}{3}\pi ∨x=1\frac{11}{12}\pi \)

(Balansmethode)
\(\sin(3x-\frac{1}{6}\pi )=-\frac{1}{2}\text{.}\)

(Exacte waardencirkel)
\(3x-\frac{1}{6}\pi =-\frac{1}{6}\pi +k⋅2\pi ∨3x-\frac{1}{6}\pi =-\frac{5}{6}\pi +k⋅2\pi \)

\(3x=k⋅2\pi ∨3x=-\frac{2}{3}\pi +k⋅2\pi \)
\(x=k⋅\frac{2}{3}\pi ∨x=-\frac{2}{9}\pi +k⋅\frac{2}{3}\pi \)

\(x\) in \([0, 2\pi ]\) geeft \(x=0∨x=\frac{2}{3}\pi ∨x=1\frac{1}{3}\pi ∨x=2\pi ∨x=\frac{4}{9}\pi ∨x=1\frac{1}{9}\pi ∨x=1\frac{7}{9}\pi \)

(Balansmethode)
\(\sin(\frac{2}{3}x+\frac{1}{4}\pi )=\frac{1}{2}\sqrt{2}\text{.}\)

(Exacte waardencirkel)
\(\frac{2}{3}x+\frac{1}{4}\pi =\frac{1}{4}\pi +k⋅2\pi ∨\frac{2}{3}x+\frac{1}{4}\pi =\frac{3}{4}\pi +k⋅2\pi \)

\(\frac{2}{3}x=k⋅2\pi ∨\frac{2}{3}x=\frac{1}{2}\pi +k⋅2\pi \)
\(x=k⋅3\pi ∨x=\frac{3}{4}\pi +k⋅3\pi \)

\(x\) in \([0, 2\pi ]\) geeft \(x=0∨x=\frac{3}{4}\pi \)

(Balansmethode)
\(\cos(2x-\frac{1}{3}\pi )=\frac{1}{2}\sqrt{3}\text{.}\)

(Exacte waardencirkel)
\(2x-\frac{1}{3}\pi =\frac{1}{6}\pi +k⋅2\pi ∨2x-\frac{1}{3}\pi =-\frac{1}{6}\pi +k⋅2\pi \)

\(2x=\frac{1}{2}\pi +k⋅2\pi ∨2x=\frac{1}{6}\pi +k⋅2\pi \)
\(x=\frac{1}{4}\pi +k⋅\pi ∨x=\frac{1}{12}\pi +k⋅\pi \)

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{4}\pi ∨x=1\frac{1}{4}\pi ∨x=\frac{1}{12}\pi ∨x=1\frac{1}{12}\pi \)

(Balansmethode)
\(4\cos(3\pi x)=-4\) dus \(\cos(3\pi x)=-1\text{.}\)

(Exacte waardencirkel)
\(3\pi x=\pi +k⋅2\pi \)

\(3\pi x=\pi +k⋅2\pi \)
\(x=\frac{1}{3}+k⋅\frac{2}{3}\)

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{3}∨x=1∨x=1\frac{2}{3}∨x=2\frac{1}{3}∨x=3∨x=3\frac{2}{3}∨x=4\frac{1}{3}∨x=5∨x=5\frac{2}{3}\)

\(\sin(2x+\frac{2}{3}\pi )=1∨\sin(2x+\frac{2}{3}\pi )=-1\)

De exacte waardencirkel geeft
\(2x+\frac{2}{3}\pi =\frac{1}{2}\pi +k⋅2\pi ∨2x+\frac{2}{3}\pi =1\frac{1}{2}\pi +k⋅2\pi \)

\(2x=-\frac{1}{6}\pi +k⋅2\pi ∨2x=\frac{5}{6}\pi +k⋅2\pi \)
\(x=-\frac{1}{12}\pi +k⋅\pi ∨x=\frac{5}{12}\pi +k⋅\pi \)

\(u=\cos(x)\) geeft \(4u^2-1\)

\(4u^2=1\)
\(u^2=\frac{1}{4}\)
\(u=-\frac{1}{2}∨u=\frac{1}{2}\)

\(\cos(x)=-\frac{1}{2}∨\cos(x)=\frac{1}{2}\) geeft
\(x=\frac{2}{3}\pi +k⋅2\pi ∨x=-\frac{2}{3}\pi +k⋅2\pi ∨x=\frac{1}{3}\pi +k⋅2\pi ∨x=-\frac{1}{3}\pi +k⋅2\pi \)

\(u=\sin(x)\) geeft \(2u^2+u\)

\(u(2u+1)=0\)
\(u=0∨2u=-1\)
\(u=0∨u=-\frac{1}{2}\)

\(\sin(x)=0∨\sin(x)=-\frac{1}{2}\) geeft
\(x=k⋅\pi ∨x=-\frac{1}{6}\pi +k⋅2\pi ∨x=-\frac{5}{6}\pi +k⋅2\pi \)

(\(\sin^2(A)+\cos^2(A)=1\) geeft)
\(2(1-\cos^2(x))+\cos^2(x)-1\frac{1}{2}\cos(x)-2\frac{1}{2}=0\)
ofwel
\(\cos^2(x)+1\frac{1}{2}\cos(x)+\frac{1}{2}=0\)

\(u=\cos(x)\) geeft \(u^2+1\frac{1}{2}u+\frac{1}{2}=0\)

\(2u^2+3u+1=0\)
\(D=3^2-4⋅2⋅1=1\)
\(u={-3-\sqrt{1} \over 2⋅2}∨u={-3+\sqrt{1} \over 2⋅2}\)
\(u=-1∨u=-\frac{1}{2}\)

\(\cos(x)=-1∨\cos(x)=-\frac{1}{2}\) geeft
\(x=\pi +k⋅2\pi ∨x=\frac{2}{3}\pi +k⋅2\pi ∨x=-\frac{2}{3}\pi +k⋅2\pi \)

\(\sin(3x-\frac{1}{2}\pi )=0∨\sin(1\frac{1}{2}x+\frac{1}{2}\pi )=0\)

(Exacte waardencirkel)
\(3x-\frac{1}{2}\pi =k⋅\pi ∨1\frac{1}{2}x+\frac{1}{2}\pi =k⋅\pi \)

\(3x=\frac{1}{2}\pi +k⋅\pi ∨1\frac{1}{2}x=-\frac{1}{2}\pi +k⋅\pi \)
\(x=\frac{1}{6}\pi +k⋅\frac{1}{3}\pi ∨x=-\frac{1}{3}\pi +k⋅\frac{2}{3}\pi \)

(Eenheidscirkel)
\(x-\frac{2}{3}\pi =2x-\frac{2}{3}\pi +k⋅2\pi ∨x-\frac{2}{3}\pi =\pi -(2x-\frac{2}{3}\pi )+k⋅2\pi \)

\(x-\frac{2}{3}\pi =2x-\frac{2}{3}\pi +k⋅2\pi ∨x-\frac{2}{3}\pi =-2x+1\frac{2}{3}\pi +k⋅2\pi \)
\(-x=k⋅2\pi ∨3x=2\frac{1}{3}\pi +k⋅2\pi \)
\(x=k⋅2\pi ∨x=\frac{7}{9}\pi +k⋅\frac{2}{3}\pi \)

(Eenheidscirkel)
\(2x+\frac{1}{4}\pi =3x-\frac{2}{3}\pi +k⋅2\pi ∨2x+\frac{1}{4}\pi =-(3x-\frac{2}{3}\pi )+k⋅2\pi \)

\(2x+\frac{1}{4}\pi =3x-\frac{2}{3}\pi +k⋅2\pi ∨2x+\frac{1}{4}\pi =-3x+\frac{2}{3}\pi +k⋅2\pi \)
\(-x=-\frac{11}{12}\pi +k⋅2\pi ∨5x=\frac{5}{12}\pi +k⋅2\pi \)
\(x=\frac{11}{12}\pi +k⋅2\pi ∨x=\frac{1}{12}\pi +k⋅\frac{2}{5}\pi \)

(Eenheidscirkel)
\(2\pi x=\pi x+\frac{3}{4}\pi +k⋅\pi \)

\(\pi x=\frac{3}{4}\pi +k⋅\pi \)
\(x=\frac{3}{4}+k⋅1\)

(\(-\cos(A)=\cos(A+\pi )\) geeft)
\(\cos(3x+\frac{1}{4}\pi )=\cos(x-\frac{3}{4}\pi +\pi )\)
\(\cos(3x+\frac{1}{4}\pi )=\cos(x+\frac{1}{4}\pi )\)

(Eenheidscirkel)
\(3x+\frac{1}{4}\pi =x+\frac{1}{4}\pi +k⋅2\pi ∨3x+\frac{1}{4}\pi =-(x+\frac{1}{4}\pi )+k⋅2\pi \)

\(3x+\frac{1}{4}\pi =x+\frac{1}{4}\pi +k⋅2\pi ∨3x+\frac{1}{4}\pi =-x-\frac{1}{4}\pi +k⋅2\pi \)
\(2x=k⋅2\pi ∨4x=-\frac{1}{2}\pi +k⋅2\pi \)
\(x=k⋅\pi ∨x=-\frac{1}{8}\pi +k⋅\frac{1}{2}\pi \)

(\(\cos(A)=\sin(A+\frac{1}{2}\pi )\) geeft)
\(\sin(2x-\frac{1}{3})=\sin(x-\frac{1}{2}\pi +\frac{1}{2}\pi )\)
\(\sin(2x-\frac{1}{3})=\sin(x)\)

(Eenheidscirkel)
\(2x-\frac{1}{3}=x+k⋅2\pi ∨2x-\frac{1}{3}=\pi -x+k⋅2\pi \)

\(2x-\frac{1}{3}=x+k⋅2\pi ∨2x-\frac{1}{3}=-x+\pi +k⋅2\pi \)
\(x=\frac{1}{3}+k⋅2\pi ∨3x=\pi +\frac{1}{3}+k⋅2\pi \)
\(x=\frac{1}{3}+k⋅2\pi ∨x=\frac{1}{3}\pi +\frac{1}{9}+k⋅\frac{2}{3}\pi \)

(\(-\cos(A)=\cos(A+\pi )\) geeft)
\(\sin(2x)=\cos(3x-1\frac{1}{2}\pi +\pi )\)
\(\sin(2x)=\cos(3x-\frac{1}{2}\pi )\)

(\(\cos(A)=\sin(A+\frac{1}{2}\pi )\) geeft)
\(\sin(2x)=\sin(3x-\frac{1}{2}\pi +\frac{1}{2}\pi )\)
\(\sin(2x)=\sin(3x)\)

(Eenheidscirkel)
\(2x=3x+k⋅2\pi ∨2x=\pi -3x+k⋅2\pi \)

\(2x=3x+k⋅2\pi ∨2x=-3x+\pi +k⋅2\pi \)
\(-x=k⋅2\pi ∨5x=\pi +k⋅2\pi \)
\(x=k⋅2\pi ∨x=\frac{1}{5}\pi +k⋅\frac{2}{5}\pi \)