Goniometrische vergelijkingen

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

\(2x=-\frac{3}{5}\pi +k⋅\pi \)
\(x=-\frac{3}{10}\pi +k⋅\frac{1}{2}\pi \)

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

(Balansmethode)
\(\sin(\frac{3}{5}x-\frac{5}{6}\pi )=-\frac{1}{2}\text{.}\)

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

\(\frac{3}{5}x=\frac{2}{3}\pi +k⋅2\pi ∨\frac{3}{5}x=k⋅2\pi \)
\(x=1\frac{1}{9}\pi +k⋅3\frac{1}{3}\pi ∨x=k⋅3\frac{1}{3}\pi \)

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

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

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

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

\(x\) in \([0, 2\pi ]\) geeft \(x=2\pi ∨x=0\)

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

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

\(\frac{1}{4}x=\frac{5}{12}\pi +k⋅2\pi ∨\frac{1}{4}x=\frac{1}{12}\pi +k⋅2\pi \)
\(x=1\frac{2}{3}\pi +k⋅8\pi ∨x=\frac{1}{3}\pi +k⋅8\pi \)

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

(Balansmethode)
\(3\sin(3x+\frac{5}{6}\pi )=-3\) dus \(\sin(3x+\frac{5}{6}\pi )=-1\text{.}\)

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

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

\(x\) in \([0, 2\pi ]\) geeft \(x=\frac{2}{9}\pi ∨x=\frac{8}{9}\pi ∨x=1\frac{5}{9}\pi \)

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

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

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

\(u=\sin(x)\) geeft \(4u^2=3\)

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

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

\(u=\cos(x)\) geeft \(u^2=u\)

\(u^2-u=0\)
\(u(u-1)=0\)
\(u=0∨u=1\)

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

(\(\sin^2(A)+\cos^2(A)=1\) geeft)
\(2(1-\sin^2(x))-2\sin^2(x)-4\sin(x)=3\)
ofwel
\(4\sin^2(x)+4\sin(x)+1=0\)

\(u=\sin(x)\) geeft \(4u^2+4u+1=0\)

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

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

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

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

\(1\frac{1}{2}x=\frac{1}{10}\pi +k⋅\pi ∨\frac{4}{5}x=\frac{1}{5}\pi +k⋅\pi \)
\(x=\frac{1}{15}\pi +k⋅\frac{2}{3}\pi ∨x=\frac{1}{4}\pi +k⋅1\frac{1}{4}\pi \)

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

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

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

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

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

\(x=k⋅\pi \)

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

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

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

(\(\cos(A)=\sin(A+\frac{1}{2}\pi )\) geeft)
\(\sin(\pi x-\frac{1}{3}\pi )=\sin(2\pi x-\frac{5}{6}\pi +\frac{1}{2}\pi )\)
\(\sin(\pi x-\frac{1}{3}\pi )=\sin(2\pi x-\frac{1}{3}\pi )\)

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

\(\pi x-\frac{1}{3}\pi =2\pi x-\frac{1}{3}\pi +k⋅2\pi ∨\pi x-\frac{1}{3}\pi =-2\pi x+1\frac{1}{3}\pi +k⋅2\pi \)
\(-\pi x=k⋅2\pi ∨3\pi x=1\frac{2}{3}\pi +k⋅2\pi \)
\(x=k⋅2∨x=\frac{5}{9}+k⋅\frac{2}{3}\)

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

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

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

\(x-\frac{3}{4}\pi =2x-\frac{3}{4}\pi +k⋅2\pi ∨x-\frac{3}{4}\pi =-2x+1\frac{3}{4}\pi +k⋅2\pi \)
\(-x=k⋅2\pi ∨3x=2\frac{1}{2}\pi +k⋅2\pi \)
\(x=k⋅2\pi ∨x=\frac{5}{6}\pi +k⋅\frac{2}{3}\pi \)