Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
3p
\(\cos(\frac{3}{4} x + \frac{1}{6} \pi ) = 0\)
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(Exacte waardencirkel)
\(\frac{3}{4} x + \frac{1}{6} \pi = \frac{1}{2} \pi + k ⋅ \pi \)
1p
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\(\frac{3}{4} x = \frac{1}{3} \pi + k ⋅ \pi \)
\(x = \frac{4}{9} \pi + k ⋅ 1\frac{1}{3} \pi \)
1p
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\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{4}{9} \pi ∨ x = 1\frac{7}{9} \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(-2 \cos(2 x - \frac{1}{3} \pi ) = -1\)
○
(Balansmethode)
\(\cos(2 x - \frac{1}{3} \pi ) = \frac{1}{2} \text{.}\)
1p
○
(Exacte waardencirkel)
\(2 x - \frac{1}{3} \pi = \frac{1}{3} \pi + k ⋅ 2 \pi ∨ 2 x - \frac{1}{3} \pi = -\frac{1}{3} \pi + k ⋅ 2 \pi \)
1p
○
\(2 x = \frac{2}{3} \pi + k ⋅ 2 \pi ∨ 2 x = k ⋅ 2 \pi \)
\(x = \frac{1}{3} \pi + k ⋅ \pi ∨ x = k ⋅ \pi \)
1p
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\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{1}{3} \pi ∨ x = 1\frac{1}{3} \pi ∨ x = 0 ∨ x = \pi ∨ x = 2 \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(3 \cos(1\frac{1}{2} x) = -1\frac{1}{2} \sqrt{2}\)
○
(Balansmethode)
\(\cos(1\frac{1}{2} x) = -\frac{1}{2} \sqrt{2} \text{.}\)
1p
○
(Exacte waardencirkel)
\(1\frac{1}{2} x = \frac{3}{4} \pi + k ⋅ 2 \pi ∨ 1\frac{1}{2} x = 1\frac{1}{4} \pi + k ⋅ 2 \pi \)
1p
○
\(1\frac{1}{2} x = \frac{3}{4} \pi + k ⋅ 2 \pi ∨ 1\frac{1}{2} x = 1\frac{1}{4} \pi + k ⋅ 2 \pi \)
\(x = \frac{1}{2} \pi + k ⋅ 1\frac{1}{3} \pi ∨ x = \frac{5}{6} \pi + k ⋅ 1\frac{1}{3} \pi \)
1p
○
\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{1}{2} \pi ∨ x = 1\frac{5}{6} \pi ∨ x = \frac{5}{6} \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(-5 \sin(2 x + \frac{1}{2} \pi ) = -2\frac{1}{2} \sqrt{3}\)
○
(Balansmethode)
\(\sin(2 x + \frac{1}{2} \pi ) = \frac{1}{2} \sqrt{3} \text{.}\)
1p
○
(Exacte waardencirkel)
\(2 x + \frac{1}{2} \pi = \frac{1}{3} \pi + k ⋅ 2 \pi ∨ 2 x + \frac{1}{2} \pi = \frac{2}{3} \pi + k ⋅ 2 \pi \)
1p
○
\(2 x = -\frac{1}{6} \pi + k ⋅ 2 \pi ∨ 2 x = \frac{1}{6} \pi + k ⋅ 2 \pi \)
\(x = -\frac{1}{12} \pi + k ⋅ \pi ∨ x = \frac{1}{12} \pi + k ⋅ \pi \)
1p
○
\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{11}{12} \pi ∨ x = 1\frac{11}{12} \pi ∨ x = \frac{1}{12} \pi ∨ x = 1\frac{1}{12} \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(2 + 4 \sin(3 x + \frac{5}{6} \pi ) = 6\)
○
(Balansmethode)
\(4 \sin(3 x + \frac{5}{6} \pi ) = 4\) dus \(\sin(3 x + \frac{5}{6} \pi ) = 1 \text{.}\)
1p
○
(Exacte waardencirkel)
\(3 x + \frac{5}{6} \pi = \frac{1}{2} \pi + k ⋅ 2 \pi \)
1p
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\(3 x = -\frac{1}{3} \pi + k ⋅ 2 \pi \)
\(x = -\frac{1}{9} \pi + k ⋅ \frac{2}{3} \pi \)
1p
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\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{5}{9} \pi ∨ x = 1\frac{2}{9} \pi ∨ x = 1\frac{8}{9} \pi \)
1p
Los exact op.
3p
\(\sin^{2}(2 x - \frac{1}{3} \pi ) = 1\)
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\(\sin(2 x - \frac{1}{3} \pi ) = 1 ∨ \sin(2 x - \frac{1}{3} \pi ) = -1\)
1p
○
De exacte waardencirkel geeft
\(2 x - \frac{1}{3} \pi = \frac{1}{2} \pi + k ⋅ 2 \pi ∨ 2 x - \frac{1}{3} \pi = 1\frac{1}{2} \pi + k ⋅ 2 \pi \)
1p
○
\(2 x = \frac{5}{6} \pi + k ⋅ 2 \pi ∨ 2 x = 1\frac{5}{6} \pi + k ⋅ 2 \pi \)
\(x = \frac{5}{12} \pi + k ⋅ \pi ∨ x = \frac{11}{12} \pi + k ⋅ \pi \)
1p
Los exact op.
3p
\(2 \cos^{2}(x) + -1\frac{1}{2} = 0\)
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\(u = \cos(x)\) geeft \(2 u^{2} + -1\frac{1}{2}\)
1p
○
\(4 u^{2} - 3 = 0\)
\(4 u^{2} = 3\)
\(u^{2} = \frac{3}{4}\)
\(u = -{1 \over 2} \sqrt{3} ∨ u = {1 \over 2} \sqrt{3}\)
1p
○
\(\cos(x) = -{1 \over 2} \sqrt{3} ∨ \cos(x) = {1 \over 2} \sqrt{3}\) geeft
\(x = {5 \over 6} \pi+k \cdot 2 \pi ∨ x = -{5 \over 6} \pi+k \cdot 2 \pi ∨ x = {1 \over 6} \pi+k \cdot 2 \pi ∨ x = -{1 \over 6} \pi+k \cdot 2 \pi\)
1p
Los exact op.
3p
\(\cos^{2}(x) = -\frac{1}{2} \cos(x)\)
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\(u = \cos(x)\) geeft \(u^{2} = -\frac{1}{2} u\)
1p
○
\(u^{2} + \frac{1}{2} u = 0\)
\(2 u^{2} + u = 0\)
\(u (2 u + 1) = 0\)
\(u = 0 ∨ 2 u = -1\)
\(u = 0 ∨ u = -\frac{1}{2}\)
1p
○
\(\cos(x) = 0 ∨ \cos(x) = -\frac{1}{2}\) geeft
\(x = {1 \over 2} \pi+k \pi ∨ x = {2 \over 3} \pi+k \cdot 2 \pi ∨ x = -{2 \over 3} \pi+k \cdot 2 \pi\)
1p
Los exact op.
4p
\(2 \cos^{2}(x) - 3 \sin(x) = 3\)
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(\(\sin^{2}(A) + \cos^{2}(A) = 1\) geeft)
\(2 (1 - \sin^{2}(x)) - 3 \sin(x) = 3\)
ofwel
\(2 \sin^{2}(x) + 3 \sin(x) + 1 = 0\)
1p
○
\(u = \sin(x)\) geeft \(2 u^{2} + 3 u + 1 = 0\)
1p
○
\(D = 3^{2} - 4 ⋅ 2 ⋅ 1 = 1\)
\(u = {-3 - \sqrt{1} \over 2 ⋅ 2} ∨ u = {-3 + \sqrt{1} \over 2 ⋅ 2}\)
\(u = -1 ∨ u = -\frac{1}{2}\)
1p
○
\(\sin(x) = -1 ∨ \sin(x) = -\frac{1}{2}\) geeft
\(x = -{1 \over 2} \pi+k \cdot 2 \pi ∨ x = -{1 \over 6} \pi+k \cdot 2 \pi ∨ x = -{5 \over 6} \pi+k \cdot 2 \pi\)
1p
Los exact op.
3p
\(1\frac{1}{3} \cos(1\frac{1}{2} x - \frac{2}{5} \pi ) \cos(\frac{4}{5} x - \frac{3}{5} \pi ) = 0\)
○
\(\cos(1\frac{1}{2} x - \frac{2}{5} \pi ) = 0 ∨ \cos(\frac{4}{5} x - \frac{3}{5} \pi ) = 0\)
1p
○
(Exacte waardencirkel)
\(1\frac{1}{2} x - \frac{2}{5} \pi = \frac{1}{2} \pi + k ⋅ \pi ∨ \frac{4}{5} x - \frac{3}{5} \pi = \frac{1}{2} \pi + k ⋅ \pi \)
1p
○
\(1\frac{1}{2} x = \frac{9}{10} \pi + k ⋅ \pi ∨ \frac{4}{5} x = 1\frac{1}{10} \pi + k ⋅ \pi \)
\(x = \frac{3}{5} \pi + k ⋅ \frac{2}{3} \pi ∨ x = 1\frac{3}{8} \pi + k ⋅ 1\frac{1}{4} \pi \)
1p
Los exact op.
3p
\(\sin(3 x-{1 \over 4} \pi) = \sin(0)\)
○
(Eenheidscirkel)
\(3 x-{1 \over 4} \pi = 0+k \cdot 2 \pi ∨ 3 x-{1 \over 4} \pi = \pi-0+k \cdot 2 \pi\)
1p
○
\(3 x-{1 \over 4} \pi = 0+k \cdot 2 \pi ∨ 3 x-{1 \over 4} \pi = \pi+k \cdot 2 \pi\)
\(3 x = {1 \over 4} \pi+k \cdot 2 \pi ∨ 3 x = 1\frac{1}{4} \pi+k \cdot 2 \pi\)
\(x = {1 \over 12} \pi+k \cdot {2 \over 3} \pi ∨ x = {5 \over 12} \pi+k \cdot {2 \over 3} \pi\)
2p
Los exact op.
3p
\(\cos(\pi x-{1 \over 3} \pi) = \cos(2 \pi x+{2 \over 3} \pi)\)
○
(Eenheidscirkel)
\(\pi x-{1 \over 3} \pi = (2 \pi x+{2 \over 3} \pi)+k \cdot 2 \pi ∨ \pi x-{1 \over 3} \pi = -(2 \pi x+{2 \over 3} \pi)+k \cdot 2 \pi\)
1p
○
\(\pi x-{1 \over 3} \pi = (2 \pi x+{2 \over 3} \pi)+k \cdot 2 \pi ∨ \pi x-{1 \over 3} \pi = (-2 \pi x-{2 \over 3} \pi)+k \cdot 2 \pi\)
\(-\pi x = \pi+k \cdot 2 \pi ∨ 3 \pi x = -{1 \over 3} \pi+k \cdot 2 \pi\)
\(x = -1+k \cdot 2 ∨ x = -{1 \over 9}+k \cdot {2 \over 3}\)
2p
Los exact op.
3p
\(\tan(x-{2 \over 3} \pi) = \tan(2 x-{1 \over 2} \pi)\)
○
(Eenheidscirkel)
\(x-{2 \over 3} \pi = (2 x-{1 \over 2} \pi)+k \pi\)
1p
○
\(-x = {1 \over 6} \pi+k \pi\)
\(x = -{1 \over 6} \pi+k \pi\)
2p
Los exact op.
4p
\(\cos(2 x) = -\cos(-1\frac{5}{6} \pi)\)
○
(\(-\cos(A) = \cos(A + \pi )\) geeft)
\(\cos(2 x) = \cos(-1\frac{5}{6} \pi+\pi)\)
\(\cos(2 x) = \cos(-{5 \over 6} \pi)\)
1p
○
(Eenheidscirkel)
\(2 x = -{5 \over 6} \pi+k \cdot 2 \pi ∨ 2 x = {5 \over 6} \pi+k \cdot 2 \pi\)
1p
○
\(2 x = -{5 \over 6} \pi+k \cdot 2 \pi ∨ 2 x = {5 \over 6} \pi+k \cdot 2 \pi\)
\(x = -{5 \over 12} \pi+k \pi ∨ x = {5 \over 12} \pi+k \pi\)
2p
Los exact op.
4p
\(\sin(x+{1 \over 3} \pi) = \cos(2 x)\)
○
(\(\cos(A) = \sin(A + {1 \over 2} \pi)\) geeft)
\(\sin(x+{1 \over 3} \pi) = \sin(2 x+{1 \over 2} \pi)\)
\(\sin(x+{1 \over 3} \pi) = \sin(2 x+{1 \over 2} \pi)\)
1p
○
(Eenheidscirkel)
\(x+{1 \over 3} \pi = (2 x+{1 \over 2} \pi)+k \cdot 2 \pi ∨ x+{1 \over 3} \pi = \pi-(2 x+{1 \over 2} \pi)+k \cdot 2 \pi\)
1p
○
\(x+{1 \over 3} \pi = (2 x+{1 \over 2} \pi)+k \cdot 2 \pi ∨ x+{1 \over 3} \pi = (-2 x+{1 \over 2} \pi)+k \cdot 2 \pi\)
\(-x = {1 \over 6} \pi+k \cdot 2 \pi ∨ 3 x = {1 \over 6} \pi+k \cdot 2 \pi\)
\(x = -{1 \over 6} \pi+k \cdot 2 \pi ∨ x = {1 \over 18} \pi+k \cdot {2 \over 3} \pi\)
2p
Los exact op.
5p
\(\sin(2 \pi x) = -\cos(\pi x-1\frac{5}{6} \pi)\)
○
(\(-\cos(A) = \cos(A + \pi )\) geeft)
\(\sin(2 \pi x) = \cos(\pi x-1\frac{5}{6} \pi+\pi)\)
\(\sin(2 \pi x) = \cos(\pi x-{5 \over 6} \pi)\)
1p
○
(\(\cos(A) = \sin(A + {1 \over 2} \pi)\) geeft)
\(\sin(2 \pi x) = \sin(\pi x-{5 \over 6} \pi+{1 \over 2} \pi)\)
\(\sin(2 \pi x) = \sin(\pi x-{1 \over 3} \pi)\)
1p
○
(Eenheidscirkel)
\(2 \pi x = (\pi x-{1 \over 3} \pi)+k \cdot 2 \pi ∨ 2 \pi x = \pi-(\pi x-{1 \over 3} \pi)+k \cdot 2 \pi\)
1p
○
\(2 \pi x = (\pi x-{1 \over 3} \pi)+k \cdot 2 \pi ∨ 2 \pi x = (-\pi x+1\frac{1}{3} \pi)+k \cdot 2 \pi\)
\(\pi x = -{1 \over 3} \pi+k \cdot 2 \pi ∨ 3 \pi x = 1\frac{1}{3} \pi+k \cdot 2 \pi\)
\(x = -{1 \over 3}+k \cdot 2 ∨ x = {4 \over 9}+k \cdot {2 \over 3}\)
2p