SOLUTION: решить уравнения: 3sinˆx+sinx*cosx=2cosˆx 4sinx-sin2x=3

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Question 824066: решить уравнения:
3sinˆx+sinx*cosx=2cosˆx
4sinx-sin2x=3

Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3sin(x) + sin(x)*cos(x) = 2cos(x)
----
3sin(x) + sin(x)cos(x) -2cos(x) = 0
---
3sin(x) + 3sin(x)cos(x)-2sin(x)cos(x) - 2cos(x) = 0
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Rearrange:
3sin(x)-2sin(x)cos(x) + 3sin(x)cos(x)-2cos(x) = 0
----
Factor:
sin(x)[3-2cos(x)] + cos(x(3-2cos(x) = 0
----
(3-2cos(x)(sin(x)+cos(x) = 0
----
cos(x) = 3/2 (that is extraneous)
----
sin(x) = -cos(x)
x = 135 degrees or (3/4)pi radians
======================================
4sin(x)-sin(2x) = 3
-----
Graphing I get x = 67.73.. degrees = 1.182.. radians
===============
Cheers,
Stan H.


Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
             3sin²(x) + sin(x)cos(x) = 2cos²(x),    0° < x < 360°

                  Вычтите 2cos²(x) с обеих сторон. Получите ноль справа.

  3sin²(x) + sin(x)cos(x) - 2cos²(x) = 0

                  Фактор как ab-2b² + 3a² = (a+b)(3a-2b)

[sin(x) + cos(x)][3sin(x) - 2cos(x)] = 0

                  Используйте свойство zero фактор

sin(x) + cos(x) = 0         3sin(x) - 2cos(x) = 0

                  Разделите каждый термин cos(x)

sin%28x%29%2Fcos%28x%29%22%22%2B%22%221%22%22=%22%22%220%22        3sin%28x%29%2Fcos%28x%29%22%22-%22%222%22%22=%22%22%220%22

    tan(x) = -1                3·tan(x) = 2
         x = 135°,315°           tan(x) = 2%2F3
                                      x = 33.69°, 326.31°

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       4sin(x) - sin²(x) = 3          0° < x < 360°
      -sin²(x) + 4sin(x) = 3
  -sin²(x) + 4sin(x) - 3 = 0
   sin²(x) - 4sin(x) + 3 = 0
[sin(x) - 1][sin(x) - 3] = 0

sin(x) - 1 = 0;    sin(x) - 3 = 0
    sin(x) = 1;        sin(x) = 3
         x = 90°;      нет решения


Есть только одно решение:  90°

Edwin