Goniometrische vergelijkingen

(Exacte waardencirkel)
\(3x+\frac{4}{5}\pi =k⋅\pi \)

\(3x=-\frac{4}{5}\pi +k⋅\pi \)
\(x=-\frac{4}{15}\pi +k⋅\frac{1}{3}\pi \)

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{15}\pi ∨x=\frac{2}{5}\pi ∨x=\frac{11}{15}\pi ∨x=1\frac{1}{15}\pi ∨x=1\frac{2}{5}\pi ∨x=1\frac{11}{15}\pi \)

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

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

\(1\frac{1}{2}x=-\frac{1}{2}\pi +k⋅2\pi ∨1\frac{1}{2}x=\frac{1}{6}\pi +k⋅2\pi \)
\(x=-\frac{1}{3}\pi +k⋅1\frac{1}{3}\pi ∨x=\frac{1}{9}\pi +k⋅1\frac{1}{3}\pi \)

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

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

(Exacte waardencirkel)
\(\frac{2}{3}\pi x-\frac{1}{6}\pi =\frac{3}{4}\pi +k⋅2\pi ∨\frac{2}{3}\pi x-\frac{1}{6}\pi =1\frac{1}{4}\pi +k⋅2\pi \)

\(\frac{2}{3}\pi x=\frac{11}{12}\pi +k⋅2\pi ∨\frac{2}{3}\pi x=1\frac{5}{12}\pi +k⋅2\pi \)
\(x=1\frac{3}{8}+k⋅3∨x=2\frac{1}{8}+k⋅3\)

\(x\) in \([0, 2\pi ]\) geeft \(x=1\frac{3}{8}∨x=4\frac{3}{8}∨x=2\frac{1}{8}∨x=5\frac{1}{8}\)

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

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

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

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

(Balansmethode)
\(5\cos(2x-\frac{3}{4}\pi )=5\) dus \(\cos(2x-\frac{3}{4}\pi )=1\text{.}\)

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

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

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{3}{8}\pi ∨x=1\frac{3}{8}\pi \)

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

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

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

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

\(4u^2-1=0\)
\(4u^2=1\)
\(u^2=\frac{1}{4}\)
\(u=-\frac{1}{2}∨u=\frac{1}{2}\)

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

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

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

\(\cos(x)=0∨\cos(x)=-\frac{1}{2}\) geeft
\(x=\frac{1}{2}\pi +k⋅\pi ∨x=\frac{2}{3}\pi +k⋅2\pi ∨x=-\frac{2}{3}\pi +k⋅2\pi \)

(\(\sin^2(A)+\cos^2(A)=1\) geeft)
\(3(1-\cos^2(x))+2\cos^2(x)=-\frac{1}{2}\cos(x)+2\frac{1}{2}\)
ofwel
\(\cos^2(x)-\frac{1}{2}\cos(x)-\frac{1}{2}=0\)

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

\(2u^2-u-1=0\)
\(D=(-1)^2-4⋅2⋅-1=9\)
\(u={1-\sqrt{9} \over 2⋅2}∨u={1+\sqrt{9} \over 2⋅2}\)
\(u=-\frac{1}{2}∨u=1\)

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

\(\cos(3x+\frac{1}{4}\pi )=0∨\cos(\frac{4}{5}x-\frac{1}{2}\pi )=0\)

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

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

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

\(x-\frac{2}{3}\pi =3x-\frac{1}{3}\pi +k⋅2\pi ∨x-\frac{2}{3}\pi =-3x+1\frac{1}{3}\pi +k⋅2\pi \)
\(-2x=\frac{1}{3}\pi +k⋅2\pi ∨4x=2\pi +k⋅2\pi \)
\(x=-\frac{1}{6}\pi +k⋅\pi ∨x=\frac{1}{2}\pi +k⋅\frac{1}{2}\pi \)

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

\(x-\frac{1}{4}\pi =2x+\frac{3}{4}\pi +k⋅2\pi ∨x-\frac{1}{4}\pi =-2x-\frac{3}{4}\pi +k⋅2\pi \)
\(-x=\pi +k⋅2\pi ∨3x=-\frac{1}{2}\pi +k⋅2\pi \)
\(x=-\pi +k⋅2\pi ∨x=-\frac{1}{6}\pi +k⋅\frac{2}{3}\pi \)

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

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

(\(-\cos(A)=\cos(A+\pi )\) geeft)
\(\cos(3x-\frac{1}{2}\pi )=\cos(2x-\frac{3}{4}\pi +\pi )\)
\(\cos(3x-\frac{1}{2}\pi )=\cos(2x+\frac{1}{4}\pi )\)

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

\(3x-\frac{1}{2}\pi =2x+\frac{1}{4}\pi +k⋅2\pi ∨3x-\frac{1}{2}\pi =-2x-\frac{1}{4}\pi +k⋅2\pi \)
\(x=\frac{3}{4}\pi +k⋅2\pi ∨5x=\frac{1}{4}\pi +k⋅2\pi \)
\(x=\frac{3}{4}\pi +k⋅2\pi ∨x=\frac{1}{20}\pi +k⋅\frac{2}{5}\pi \)

(\(\cos(A)=\sin(A+\frac{1}{2}\pi )\) geeft)
\(\sin(2x)=\sin(-\frac{3}{4}\pi +\frac{1}{2}\pi )\)
\(\sin(2x)=\sin(-\frac{1}{4}\pi )\)

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

\(2x=-\frac{1}{4}\pi +k⋅2\pi ∨2x=1\frac{1}{4}\pi +k⋅2\pi \)
\(x=-\frac{1}{8}\pi +k⋅\pi ∨x=\frac{5}{8}\pi +k⋅\pi \)

(\(-\cos(A)=\cos(A+\pi )\) geeft)
\(\sin(x+\frac{3}{4})=\cos(2x-\frac{3}{4}-1\frac{1}{2}\pi +\pi )\)
\(\sin(x+\frac{3}{4})=\cos(2x-\frac{3}{4}-\frac{1}{2}\pi )\)

(\(\cos(A)=\sin(A+\frac{1}{2}\pi )\) geeft)
\(\sin(x+\frac{3}{4})=\sin(2x-\frac{3}{4}-\frac{1}{2}\pi +\frac{1}{2}\pi )\)
\(\sin(x+\frac{3}{4})=\sin(2x-\frac{3}{4})\)

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

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