De som van alle oplossingen van
cos 2x + cos x = 0
gelegen in het interval [ 0, 2π [
is (in radialen)
|
A. \(\boldsymbol{\frac{13\pi}{6}}\) |
|---|
| B. \(\boldsymbol{2\pi }\) |
| C. \(\boldsymbol{3\pi}\) |
| D. \(\boldsymbol{\frac{17\pi}{6}}\) |
| E. \(\boldsymbol{\frac{7\pi}{3}}\) |
[ 5-4450 - op net sinds 6.12.09-(E)-9.12.2024 ]
Translation in E N G L I S H
The sum of all the solutions of
cos 2x + cos x = 0
in the interval [ 0, 2π)
is (in radians)
|
A. \(\boldsymbol{\frac{13\pi}{6}}\) |
|---|
| B. \(\boldsymbol{2\pi}\) |
| C. \(\boldsymbol{3\pi}\) |
| D. \(\boldsymbol{\frac{17\pi}{6}}\) |
| E. \(\boldsymbol{\frac{7\pi}{3}}\) |
Oplossing - Solution
1ste manier :
cos 2x + cos x = 0
⇔ cos 2x = − cos x
⇔ cos 2x = cos(180° − x)
⇔ 2x = 180° − x + k.360° ∨ 2x = x − 180° + k.360°
⇔ 3x = 180° + k.360° ∨ x = −180° + k.360° = 180° + k.360°
⇔ x = 60° + k.120° ∨ x = 180° + k.360°
⇔ x = 60° + k.120°
2de manier :
cos 2x + cos x = 0
⇔ 2.cos²x − 1 + cos 2x + = 0
⇔ cos x = t ∧ 2t² + t − 1 = 0
⇔ cos x = t ∧ (t + 1)(2t − 1) = 0
⇔ cos x = t ∧ [ t = −1 ∨ t = ½ ]
⇔ cos x = −1 ∨ cos x = ½
⇔ x = 180° + k.360° ∨ x = ±60° + k.360°
-----------------------------
In beide gevallen vind je 60°, 180°, 300° in het interval [0°,360°[
De som is 540° of 3π