|
© h.hofstede (h.hofstede@hogeland.nl) |
|||
 |
|||
| 1. | a. | 5 -
2tan(1 - x) = 8 -2tan(1 - x) = 3 tan(1 - x) = -1,5 1 - x = -0,98 + kπ -x = -1,98 + kπ x = 1,98 + kπ In interval [0, 2π] geeft dat de oplossingen {1.98, 5.12} |
|
| b. | 2 + tan(2x + 8) = 12 tan(2x + 8) = 10 2x + 8 = 1,47 + kπ 2x = -6,53 + kπ x = -3,26 + k • 1/2π In interval [0, 2π] geeft dat de oplossingen {1.45, 3.02, 4.59, 6.66} |
||
| c. | 4tan2x - 16 =
0 4tan2x = 16 tan2x = 4 tanx = 2 ∨ tanx = -2 x = 1,11 + kπ ∨ x = -1,11 + kπ In interval [0, 2π] geeft dat de oplossingen {1.11, 2.03, 4.25, 5.18} |
||
| d. | 3tan(0,5x) = 5
- tan(0,5x) 4tan(0,5x) = 5 tan(0,5x) = 1,25 0,5x = 0,90 + kπ x = 1,79 + k4π In interval [0, 2π] geeft dat de oplossing {1.79} |
||
| 2. | a. | 2tan2x + 1 = 7 2tan2x = 6 tan2x = 3 tanx = √3 ∨ tanx = -√3 x = 1/3π + kπ ∨ x = -1/3π + kπ In interval [0, 2π] geeft dat de oplossingen {1/3π, 2/3π, 11/3π, 12/3π} |
|
| b. | tan2x
- tanx = 0 tanx(tanx - 1) = 0 tanx = 0 ∨ tanx = 1 x = 0 + kπ ∨ x = 1/4π + kπ In interval [0, 2π] geeft dat de oplossingen {0, 1/4π, π, 11/4π} |
||
| c. | 2tan(x
- 1/2π)
= 2/3√3 tan(x - 1/2π) = 1/3√3 x - 1/2π = 1/6π + kπ x = 2/3π + kπ In interval [0, 2π] geeft dat de oplossingen {2/3π, 12/3π} |
||
| d. | tan(2x +
π) =
tan(1/3π
+ x) 2x + π = 1/3π + x + kπ x = -2/3π + kπ In interval [0, 2π] geeft dat de oplossingen {1/3π, 11/3π} |
||
| 3. | a. | tanx = 2sinx sinx/cosx = 2sinx sinx = 2sinxcosx sinx - 2sinxcosx = 0 sinx(1 - 2cosx) = 0 sinx = 0 cosx = 1/2 x = 0 + k2π ∨ x = π - 0 + k2π ∨ x = 1/3π + k2π ∨ x = 2π - 1/3π + k2π In interval [0, 2π] geeft dat de oplossingen {0, 1/3π, π, 12/3π, 2π} |
|
| b. | tan2x =
4/9
· cos2x sin²x/cos²x = 4/9·cos2x sin2(x) = 4/9 · cos4x 9(1 - cos2x) = 4cos4x 4cos4x + 9cos2x - 9 = 0 cos2x = (-9 ± Ö225)/8 cos2x = 3/4 ∨ cos2x = -3 cosx = 1/2√3 ∨ cosx = -1/2√3 In interval [0, 2π] geeft dat de oplossingen {1/6p, 5/6p, 7/6p, 11/6p} |
||
| c. | 3tanx + 2cosx = 0 3 • sinx/cosx + 2cosx = 0 3sinx + 2cos2x = 0 3sinx + 2(1 - sin2x) = 0 3sinx + 2 - 2sin2x = 0 noem sinx = p -2p2 + 3p + 2 = 0 ABC-formule: p = (-3 ±√(9 + 16))/-4 = (-3 ± 5)/-4 = 2 of -1/2 sinx = 2 kan niet, dus blijft over sinx = -1/2 x = 11/6π + k2π ∨ x = π - 11/6π + k2π In interval [0, 2π] geeft dat de oplossingen {11/6π, 15/6π} |
||
| 4. |
 |
||
| 5. |
 |
||
| 6. | (1 + tanx)/(1
+ cosx) = 0 1 + tanx = 0 (en 1 + cosx ¹ 0) tanx = -1 (en x ¹ π) x = 3/4π ∨ x = 13/4π Hiernaast staat de grafiek van y = (1 + tanx)/(1 + cosx) Dat is groter of gelijk aan nul voor; [0, 1/2π〉 en [3/4π, π〉 en 〈π, 11/2π〉 en [13/4π, 2π] |
 |
|
| 7. | tan(90º
- x) =
sin(90º - x)/cos(90º
- x) = cosx/sinx
= 1/tanx tan1° • tan2° • tan3° • ... •
tan89° |
||