Goniometrische vergelijkingen

De exacte waardencirkel geeft
\(\frac{4}{5}\pi x-\frac{3}{5}\pi =k⋅\pi \)

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

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

Balansmethode geeft \(\sin(\frac{3}{4}x+\frac{1}{6}\pi )=\frac{1}{2}\text{.}\)

De exacte waardencirkel geeft
\(\frac{3}{4}x+\frac{1}{6}\pi =\frac{1}{6}\pi +k⋅2\pi ∨\frac{3}{4}x+\frac{1}{6}\pi =\frac{5}{6}\pi +k⋅2\pi \)

\(\frac{3}{4}x=k⋅2\pi ∨\frac{3}{4}x=\frac{2}{3}\pi +k⋅2\pi \)
\(x=k⋅2\frac{2}{3}\pi ∨x=\frac{8}{9}\pi +k⋅2\frac{2}{3}\pi \)

\(x\) in \([0, 2\pi ]\) geeft \(x=0∨x=\frac{8}{9}\pi \)

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

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

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

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

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

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

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

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

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

De exacte waardencirkel geeft
\(4x+\frac{1}{5}\pi =\pi +k⋅2\pi \)

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

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{1}{5}\pi ∨x=\frac{7}{10}\pi ∨x=1\frac{1}{5}\pi ∨x=1\frac{7}{10}\pi \)

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

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

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

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

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

\(\frac{3}{4}x=\frac{1}{5}\pi +k⋅\pi ∨\frac{3}{4}x=1\frac{1}{10}\pi +k⋅\pi \)
\(x=\frac{4}{15}\pi +k⋅1\frac{1}{3}\pi ∨x=1\frac{7}{15}\pi +k⋅1\frac{1}{3}\pi \)