# i need help trying to find the real numbers in the interval 0, 2pi

#### Math3000

##### New member
How to find all the real number [0,2pi] that satisfy --->>> 2sinx + cosx + cos = 0

#### Otis

##### Senior Member
2sinx + cosx + cos = 0
That part is not correct. Did you misread the given equation?

(2sinx)(cosx) + cosx = 0

Please try again. Hint: Factor the left-hand side.

#### Math3000

##### New member
That's not correct. Did you misread the given equation?

(2sinx)(cosx) + cosx = 0

Please try again. Hint: Factor the left-hand side.

yes you right, i meant to type 2sinx + cosx + cosx = 0

Thank for the hint.

#### Otis

##### Senior Member
… i meant to type 2sinx + cosx + cosx = 0 …
You changed 'cos' to 'cosx' above (fixing a typo that I had ignored), but that plus sign is still wrong, heh.

#### Math3000

##### New member
You changed 'cos' to 'cosx' above (fixing a typo that I had ignored), but that plus sign is still wrong, heh.

omg you have some good eyes lol, correction 2sinx cosx + cosx = 0

Staff member

#### JeffM

##### Elite Member
Consider the general problem:

$$\displaystyle \text {Given } a,\ x \in \mathbb R, \ 0 < |a| \le 1,\ 0 \le x \le 2 \pi , \text { and } a * sin(x) * cos(x) + cos(x) = 0, \text { find } x.$$

Using Subhotosh's method, that general problem can be solved using the arcsin function.

$$\displaystyle a * sin(x) * cos(x) + cos(x) = 0 \implies cos(x)\{a * sin(x) + 1) = 0 \implies$$

$$\displaystyle cos(x) = 0 = \sqrt{1 - sin^2(x)} \text { or } a * sin(x) - 1 = 0 \implies sin(x) = \pm 1 \text { or } sin(x) = -\ \dfrac{1}{a}.$$

I'll bet dollars to doughnuts, however, that the specific problem is designed to be solved by special triangles. Is there any problem that requires knowledge of special triangles for solution? I don't get the pedagogic value of teaching a method that is far from general when a general method exists and takes only a decent calculator to apply.

#### Math3000

##### New member
Consider the general problem:

$$\displaystyle \text {Given } a,\ x \in \mathbb R, \ 0 < |a| \le 1,\ 0 \le x \le 2 \pi , \text { and } a * sin(x) * cos(x) + cos(x) = 0, \text { find } x.$$

Using Subhotosh's method, that general problem can be solved using the arcsin function.

$$\displaystyle a * sin(x) * cos(x) + cos(x) = 0 \implies cos(x)\{a * sin(x) + 1) = 0 \implies$$

$$\displaystyle cos(x) = 0 = \sqrt{1 - sin^2(x)} \text { or } a * sin(x) - 1 = 0 \implies sin(x) = \pm 1 \text { or } sin(x) = -\ \dfrac{1}{a}.$$

I'll bet dollars to donoughts, however, that the specific problem is designed to be solved by special triangles. Is there any problem that requires knowledge of special triangles for solution? I don't get the pedagogic value of teaching a method that is far from general when a general method exists and takes only a decent calculator to apply.
I'm going try to solve it with the arcsin function, thank!

#### Subhotosh Khan

##### Super Moderator
Staff member
Consider the general problem:
........I'll bet dollars to donoughts, however, that the specific problem is designed to be solved by special triangles. Is there any problem that requires knowledge of special triangles for solution? I don't get the pedagogic value of teaching a method that is far from general when a general method exists and takes only a decent calculator to apply.

#### Math3000

##### New member
no luck on this problem yet! smh

#### Subhotosh Khan

##### Super Moderator
Staff member
Now factor out cos(x) to get:

2sin(x)* cos(x) + cos(x) = 0

cos(x) * [2sin(x) + 1] = 0 Thus

Either

cos(x) = 0

or

2sin(x) + 1 = 0

Continue....

What answers did you get by following the hint above?

#### JeffM

##### Elite Member
no luck on this problem yet! smh
OK. You followed Subhotosh's explanation of the factoring and the use of the zero product property, right? You come up with two possibilities. Let's follow through on one of them. I am going to assume that you are working in radians.

$$\displaystyle cos(x) = 0 \implies \sqrt{cos^2(x)} = 0 \implies \sqrt{1 - sin^2(x)} = 0 \implies$$

$$\displaystyle 1 - sin^2(x) = 0 \implies sin^2(x) = 1 \implies sin(x) = \pm 1.$$

Probably you already know from the unit circle definition of the trigonometric functions a value of x such that sin(x) = 1, but suppose you do not. Then you can use the arcsin function. Set your calculator to DEGREES.

$$\displaystyle sin(x) = 1 \implies x = 90 \text { degrees.}$$

Now to get an exact answer in radians, use the conversion ratio of $$\displaystyle \dfrac{\pi}{180 \text { degrees}}.$$

$$\displaystyle \therefore x = 90 \text { degrees}* \dfrac{\pi}{180 \text { degrees}} = \dfrac{\pi}{2}.$$

When you do that process for sin(x) = -\ 1, you will get an answer of $$\displaystyle -\ \dfrac{\pi}{2}.$$

That answer is outside your desired range, but the sine and cosine functions are fully cyclic with a period of 2pi.

$$\displaystyle x = -\ \dfrac{\pi}{2} + 2 \pi = \dfrac{3 \pi}{2}.$$

You try it with $$\displaystyle 2sin(x) + 1 = 0.$$