Hello,
There is a rule that stay:
Sen^(2)x + Cos^(2)x = 1
Let's isolate Sen or Cos
Choosing Sen:
Sen^(2)x = 1 - Cos^(2)x
Moving the square root to the right side:
Senx = √(1 - Cos^(2)x )
Then, we stay with:
Senx + 2Cosx = 0
√(1 - Cos^(2)x ) + 2Cosx = 0
Making y = Cosx
√(1 - y^2) +2y = 0
Passing 2y to the right side of equation, but changing the sinal.
√(1 - y^2) = -2y
Raising both sides to the square:
| 1 - y^2| = (-2y)^2
|1 - y^2| = 4y^2
There is two Two possibilities.
1 - y^2 = 4y^2
or
1 - y^2 = - 4y^2
But, 1 - y^2 = -4y^2 Not is possible
Look:
1 = -3y^2
y^2 = - 1 / 3
y = +/- √(-1/3)
y = +/- √(-1/3) × √(3/3)
y = +/- √(-3)/√(9)
y = +/- √(-3)/3
This is complexo number
Where, i^2 = -1
y = +/- √(i^2.3)/3
y = +/- i.√(3)/3
Then you tell me, If you already calculated complex number. Okay?
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Following the example second:
1 - y^2 = 4y^2
1 = 4y^2 + y^2
1 = 5y^2
1/5 = y^2
y^2 = 1/5
y = +/- √(1/5)
Multiply √ (1/5) for √ (5/5)
y = +/- √(1/5) × √(5/5)
y = +/- √(1.5/5.5)
y = +/- √(5/25)
y = +/- √5 / √25
y = +/- √(5) / 5
Then,
y = √(5)/5
And,
y = - √(5) / 5
As Cosx = y
Then us stay with:
Cosx = √(5)/5
And
Cos(x) = - √(5)/5
Applying ArcCos on both the sides:
ArcCos(Cosx) = ArcCos(√(5)/5)
And
ArcCos(Cosx) = ArcCos(- √(5)/5)
1 Ex:
x = ArcCos( √(5)/5 )
2 Ex:
x = ArcCos( -√(5)/5)
Then, the value of X is
x ~ 63,43°
And
x ~ 116,56°
But, x = 63,43° There is an error of 1.78 °, even putting all the decimal places. Now the 116.56 ° angle with all the decinal places satisfies the equation.
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x could be too:
116,56° + 180° = 296,56°
Then,
x = 296,56° and 116,56°
Approximately
