Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
3p
\(\cos(\frac{2}{3} x - \frac{2}{3} \pi ) = 0\)
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(Exacte waardencirkel)
\(\frac{2}{3} x - \frac{2}{3} \pi = \frac{1}{2} \pi + k ⋅ \pi \)
1p
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\(\frac{2}{3} x = 1\frac{1}{6} \pi + k ⋅ \pi \)
\(x = 1\frac{3}{4} \pi + k ⋅ 1\frac{1}{2} \pi \)
1p
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\(x\) in \([0 , 2 \pi ]\) geeft \(x = 1\frac{3}{4} \pi ∨ x = \frac{1}{4} \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(-2 \cos(\frac{4}{5} x + \frac{1}{3} \pi ) = 1\)
○
(Balansmethode)
\(\cos(\frac{4}{5} x + \frac{1}{3} \pi ) = -\frac{1}{2} \text{.}\)
1p
○
(Exacte waardencirkel)
\(\frac{4}{5} x + \frac{1}{3} \pi = \frac{2}{3} \pi + k ⋅ 2 \pi ∨ \frac{4}{5} x + \frac{1}{3} \pi = -\frac{2}{3} \pi + k ⋅ 2 \pi \)
1p
○
\(\frac{4}{5} x = \frac{1}{3} \pi + k ⋅ 2 \pi ∨ \frac{4}{5} x = -\pi + k ⋅ 2 \pi \)
\(x = \frac{5}{12} \pi + k ⋅ 2\frac{1}{2} \pi ∨ x = -1\frac{1}{4} \pi + k ⋅ 2\frac{1}{2} \pi \)
1p
○
\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{5}{12} \pi ∨ x = 1\frac{1}{4} \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(-5 \sin(\frac{4}{5} x + \frac{3}{4} \pi ) = 2\frac{1}{2} \sqrt{2}\)
○
(Balansmethode)
\(\sin(\frac{4}{5} x + \frac{3}{4} \pi ) = -\frac{1}{2} \sqrt{2} \text{.}\)
1p
○
(Exacte waardencirkel)
\(\frac{4}{5} x + \frac{3}{4} \pi = 1\frac{1}{4} \pi + k ⋅ 2 \pi ∨ \frac{4}{5} x + \frac{3}{4} \pi = 1\frac{3}{4} \pi + k ⋅ 2 \pi \)
1p
○
\(\frac{4}{5} x = \frac{1}{2} \pi + k ⋅ 2 \pi ∨ \frac{4}{5} x = \pi + k ⋅ 2 \pi \)
\(x = \frac{5}{8} \pi + k ⋅ 2\frac{1}{2} \pi ∨ x = 1\frac{1}{4} \pi + k ⋅ 2\frac{1}{2} \pi \)
1p
○
\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{5}{8} \pi ∨ x = 1\frac{1}{4} \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(2 \cos(\frac{2}{3} x + \frac{1}{2} \pi ) = -\sqrt{3}\)
○
(Balansmethode)
\(\cos(\frac{2}{3} x + \frac{1}{2} \pi ) = -\frac{1}{2} \sqrt{3} \text{.}\)
1p
○
(Exacte waardencirkel)
\(\frac{2}{3} x + \frac{1}{2} \pi = \frac{5}{6} \pi + k ⋅ 2 \pi ∨ \frac{2}{3} x + \frac{1}{2} \pi = -\frac{5}{6} \pi + k ⋅ 2 \pi \)
1p
○
\(\frac{2}{3} x = \frac{1}{3} \pi + k ⋅ 2 \pi ∨ \frac{2}{3} x = -1\frac{1}{3} \pi + k ⋅ 2 \pi \)
\(x = \frac{1}{2} \pi + k ⋅ 3 \pi ∨ x = -2 \pi + k ⋅ 3 \pi \)
1p
○
\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{1}{2} \pi ∨ x = \pi \)
1p
Bereken zo mogelijk exact de oplossingen in \([0 , 2 \pi ] \text{.}\)
4p
\(-3 - 4 \cos(2 x - \frac{1}{5} \pi ) = 1\)
○
(Balansmethode)
\(-4 \cos(2 x - \frac{1}{5} \pi ) = 4\) dus \(\cos(2 x - \frac{1}{5} \pi ) = -1 \text{.}\)
1p
○
(Exacte waardencirkel)
\(2 x - \frac{1}{5} \pi = \pi + k ⋅ 2 \pi \)
1p
○
\(2 x = 1\frac{1}{5} \pi + k ⋅ 2 \pi \)
\(x = \frac{3}{5} \pi + k ⋅ \pi \)
1p
○
\(x\) in \([0 , 2 \pi ]\) geeft \(x = \frac{3}{5} \pi ∨ x = 1\frac{3}{5} \pi \)
1p
Los exact op.
3p
\(\cos^{2}(3 x + \frac{1}{3} \pi ) = 1\)
○
\(\cos(3 x + \frac{1}{3} \pi ) = 1 ∨ \cos(3 x + \frac{1}{3} \pi ) = -1\)
1p
○
De exacte waardencirkel geeft
\(3 x + \frac{1}{3} \pi = k ⋅ 2 \pi ∨ 3 x + \frac{1}{3} \pi = \pi + k ⋅ 2 \pi \)
1p
○
\(3 x = -\frac{1}{3} \pi + k ⋅ 2 \pi ∨ 3 x = \frac{2}{3} \pi + k ⋅ 2 \pi \)
\(x = -\frac{1}{9} \pi + k ⋅ \frac{2}{3} \pi ∨ x = \frac{2}{9} \pi + k ⋅ \frac{2}{3} \pi \)
1p
Los exact op.
3p
\(2 \sin^{2}(x) = 1\)
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\(u = \sin(x)\) geeft \(2 u^{2} = 1\)
1p
○
\(2 u^{2} - 1 = 0\)
\(2 u^{2} = 1\)
\(u^{2} = \frac{1}{2}\)
\(u = -{1 \over 2} \sqrt{2} ∨ u = {1 \over 2} \sqrt{2}\)
1p
○
\(\sin(x) = -{1 \over 2} \sqrt{2} ∨ \sin(x) = {1 \over 2} \sqrt{2}\) geeft
\(x = -{1 \over 4} \pi+k \cdot 2 \pi ∨ x = -{3 \over 4} \pi+k \cdot 2 \pi ∨ x = {1 \over 4} \pi+k \cdot 2 \pi ∨ x = {3 \over 4} \pi+k \cdot 2 \pi\)
1p
Los exact op.
3p
\(2 \cos^{2}(x) + 3 \cos(x) + 1 = 0\)
○
\(u = \cos(x)\) geeft \(2 u^{2} + 3 u + 1\)
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
○
\(\cos(x) = -1 ∨ \cos(x) = -\frac{1}{2}\) geeft
\(x = \pi+k \cdot 2 \pi ∨ x = {2 \over 3} \pi+k \cdot 2 \pi ∨ x = -{2 \over 3} \pi+k \cdot 2 \pi\)
1p
Los exact op.
4p
\(2 \sin^{2}(x) + 3 \cos(x) = 3\)
○
(\(\sin^{2}(A) + \cos^{2}(A) = 1\) geeft)
\(2 (1 - \cos^{2}(x)) + 3 \cos(x) = 3\)
ofwel
\(2 \cos^{2}(x) - 3 \cos(x) + 1 = 0\)
1p
○
\(u = \cos(x)\) geeft \(2 u^{2} - 3 u + 1 = 0\)
1p
○
\(D = (-3)^{2} - 4 ⋅ 2 ⋅ 1 = 1\)
\(u = {3 - \sqrt{1} \over 2 ⋅ 2} ∨ u = {3 + \sqrt{1} \over 2 ⋅ 2}\)
\(u = \frac{1}{2} ∨ u = 1\)
1p
○
\(\cos(x) = \frac{1}{2} ∨ \cos(x) = 1\) geeft
\(x = {1 \over 3} \pi+k \cdot 2 \pi ∨ x = -{1 \over 3} \pi+k \cdot 2 \pi ∨ x = k \cdot 2 \pi\)
1p
Los exact op.
3p
\(1\frac{3}{5} \sin(\frac{4}{5} x + \frac{1}{2} \pi ) \cos(4 x) = 0\)
○
\(\sin(\frac{4}{5} x + \frac{1}{2} \pi ) = 0 ∨ \cos(4 x) = 0\)
1p
○
(Exacte waardencirkel)
\(\frac{4}{5} x + \frac{1}{2} \pi = k ⋅ \pi ∨ 4 x = \frac{1}{2} \pi + k ⋅ \pi \)
1p
○
\(\frac{4}{5} x = -\frac{1}{2} \pi + k ⋅ \pi ∨ 4 x = \frac{1}{2} \pi + k ⋅ \pi \)
\(x = -\frac{5}{8} \pi + k ⋅ 1\frac{1}{4} \pi ∨ x = \frac{1}{8} \pi + k ⋅ \frac{1}{4} \pi \)
1p
Los exact op.
3p
\(\sin(3 \pi x-{1 \over 2} \pi) = \sin(2 \pi x+{1 \over 2} \pi)\)
○
(Eenheidscirkel)
\(3 \pi x-{1 \over 2} \pi = (2 \pi x+{1 \over 2} \pi)+k \cdot 2 \pi ∨ 3 \pi x-{1 \over 2} \pi = \pi-(2 \pi x+{1 \over 2} \pi)+k \cdot 2 \pi\)
1p
○
\(3 \pi x-{1 \over 2} \pi = (2 \pi x+{1 \over 2} \pi)+k \cdot 2 \pi ∨ 3 \pi x-{1 \over 2} \pi = (-2 \pi x+{1 \over 2} \pi)+k \cdot 2 \pi\)
\(\pi x = \pi+k \cdot 2 \pi ∨ 5 \pi x = \pi+k \cdot 2 \pi\)
\(x = 1+k \cdot 2 ∨ x = {1 \over 5}+k \cdot {2 \over 5}\)
2p
Los exact op.
3p
\(\cos(x) = \cos(3 x+{2 \over 3} \pi)\)
○
(Eenheidscirkel)
\(x = (3 x+{2 \over 3} \pi)+k \cdot 2 \pi ∨ x = -(3 x+{2 \over 3} \pi)+k \cdot 2 \pi\)
1p
○
\(x = (3 x+{2 \over 3} \pi)+k \cdot 2 \pi ∨ x = (-3 x-{2 \over 3} \pi)+k \cdot 2 \pi\)
\(-2 x = {2 \over 3} \pi+k \cdot 2 \pi ∨ 4 x = -{2 \over 3} \pi+k \cdot 2 \pi\)
\(x = -{1 \over 3} \pi+k \pi ∨ x = -{1 \over 6} \pi+k \cdot {1 \over 2} \pi\)
2p
Los exact op.
3p
\(\tan(\pi x-{1 \over 3} \pi) = \tan(-{1 \over 3} \pi)\)
○
(Eenheidscirkel)
\(\pi x-{1 \over 3} \pi = -{1 \over 3} \pi+k \pi\)
1p
○
\(\pi x = 0+k \pi\)
\(x = k \cdot 1\)
2p
Los exact op.
4p
\(\sin(2 x-{1 \over 4} \pi) = -\sin(3 x-{1 \over 4} \pi)\)
○
(\(-\sin(A) = \sin(A + \pi )\) geeft)
\(\sin(2 x-{1 \over 4} \pi) = \sin(3 x-{1 \over 4} \pi+\pi)\)
\(\sin(2 x-{1 \over 4} \pi) = \sin(3 x+{3 \over 4} \pi)\)
1p
○
(Eenheidscirkel)
\(2 x-{1 \over 4} \pi = (3 x+{3 \over 4} \pi)+k \cdot 2 \pi ∨ 2 x-{1 \over 4} \pi = \pi-(3 x+{3 \over 4} \pi)+k \cdot 2 \pi\)
1p
○
\(2 x-{1 \over 4} \pi = (3 x+{3 \over 4} \pi)+k \cdot 2 \pi ∨ 2 x-{1 \over 4} \pi = (-3 x+{1 \over 4} \pi)+k \cdot 2 \pi\)
\(-x = \pi+k \cdot 2 \pi ∨ 5 x = {1 \over 2} \pi+k \cdot 2 \pi\)
\(x = -\pi+k \cdot 2 \pi ∨ x = {1 \over 10} \pi+k \cdot {2 \over 5} \pi\)
2p
Los exact op.
4p
\(\sin(2 x-{1 \over 4} \pi) = \cos(x-{1 \over 2} \pi)\)
○
(\(\cos(A) = \sin(A + {1 \over 2} \pi)\) geeft)
\(\sin(2 x-{1 \over 4} \pi) = \sin(x-{1 \over 2} \pi+{1 \over 2} \pi)\)
\(\sin(2 x-{1 \over 4} \pi) = \sin(x)\)
1p
○
(Eenheidscirkel)
\(2 x-{1 \over 4} \pi = x+k \cdot 2 \pi ∨ 2 x-{1 \over 4} \pi = \pi-x+k \cdot 2 \pi\)
1p
○
\(2 x-{1 \over 4} \pi = x+k \cdot 2 \pi ∨ 2 x-{1 \over 4} \pi = (-x+\pi)+k \cdot 2 \pi\)
\(x = {1 \over 4} \pi+k \cdot 2 \pi ∨ 3 x = 1\frac{1}{4} \pi+k \cdot 2 \pi\)
\(x = {1 \over 4} \pi+k \cdot 2 \pi ∨ x = {5 \over 12} \pi+k \cdot {2 \over 3} \pi\)
2p
Los exact op.
5p
\(\cos(3 x+{1 \over 3} \pi) = -\sin(x-\pi)\)
○
(\(-\sin(A) = \sin(A + \pi )\) geeft)
\(\cos(3 x+{1 \over 3} \pi) = \sin(x-\pi+\pi)\)
\(\cos(3 x+{1 \over 3} \pi) = \sin(x)\)
1p
○
(\(\sin(A) = \cos(A + -{1 \over 2} \pi)\) geeft)
\(\cos(3 x+{1 \over 3} \pi) = \cos(x-{1 \over 2} \pi)\)
\(\cos(3 x+{1 \over 3} \pi) = \cos(x-{1 \over 2} \pi)\)
1p
○
(Eenheidscirkel)
\(3 x+{1 \over 3} \pi = (x-{1 \over 2} \pi)+k \cdot 2 \pi ∨ 3 x+{1 \over 3} \pi = -(x-{1 \over 2} \pi)+k \cdot 2 \pi\)
1p
○
\(3 x+{1 \over 3} \pi = (x-{1 \over 2} \pi)+k \cdot 2 \pi ∨ 3 x+{1 \over 3} \pi = (-x+{1 \over 2} \pi)+k \cdot 2 \pi\)
\(2 x = -{5 \over 6} \pi+k \cdot 2 \pi ∨ 4 x = {1 \over 6} \pi+k \cdot 2 \pi\)
\(x = -{5 \over 12} \pi+k \pi ∨ x = {1 \over 24} \pi+k \cdot {1 \over 2} \pi\)
2p