Sinx + 2cosx = 0 where 0< =x< =2pi, what can x be equal to?

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?


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.



x could be too:

116,56° + 180° = 296,56°

Then,

x = 296,56° and 116,56°

Approximately

Which of the following values are in the range is the function graphed below? check all that applya.3b.1c.0d.-2e.5f.2a