Where am I going wrong? cos2x + 2six = 0

[whitecanvas]

Please help me find out where I'm going wrong. I've been working this over and over, but cannot match my answers to the choices I have.

Q: Find all solutions in the interval [0, 2pi).

cos2x + 2sinx = 0

I use the double-angle identity to get:

1 - 2sin[sup:2k1kt9ov]2[/sup:2k1kt9ov]x + 2sinx = 0

And then I rearrange it so I can use it in the quadratic formula:

-2sin[sup:2k1kt9ov]2[/sup:2k1kt9ov]x + 2sinx + 1 = 0

So using it in the quadratic formula: [attachment=0:2k1kt9ov]mathhelp.jpg[/attachment:2k1kt9ov]

The two answers are: -0.366 and 1.366

But the answers I got aren't the choices available. So where exactly am I going wrong? Please help.

Thanks!

 

[Aladdin]

I know that : Cos(2x)= 2sin(x)cos(x) --> (1)

I have to solve this in a given interval : cos2x + 2sinx = 0 ---> (2)

Substitute (1) in (2) : I get : 2sinx.cosx + 2sinx = 0

Continue by taking sinx as a common factor

 

[soroban]

Hello, whitecanvas!

Find all solutions in the interval \(\displaystyle [0, 2\pi)\!:\;\;\cos2x + 2\sin x \:=\: 0\)

I use the double-angle identity to get: .\(\displaystyle 1 - 2\sin^2\!x + 2\sin x \:=\: 0\)

And then I rearrange it so I can use it in the quadratic formula: .\(\displaystyle -2\sin^2\!x + 2\sin x + 1 \:=\: 0\)
All this is correct . . . Good work!
And your next step is also correct, but . . .

\(\displaystyle \text{Quadratic Formula: }\;\sin x \;=\;\frac{-2 \pm \sqrt{2^2 - 4(-2)(1)}}{2(2)} \;=\;\frac{2\pm\sqrt{12}}{4} \quad\Rightarrow\quad \sin x\;=\;\frac{1\pm\sqrt{3}}{2}\)

.\(\displaystyle \text{Get it? You solved for }\sin x\text{, not }x.\)


\(\displaystyle \text{We have: }\;\sin x \:=\:\begin{Bmatrix} \frac{1+\sqrt{3}}{2} &=& 1.3660 & \text{impossible} \\ \frac{1-\sqrt{3}}{2} &=& -0.3660 & \text{okay!} \end{Bmatrix}\)


\(\displaystyle \text{Then: }\;\sin x \:=\:-0.3660 \quad\Rightarrow\quad x \;=\;\arcsin(-0.3660) \quad\Rightarrow\quad x \;=\;-0.3747\text{ radians}{\)

\(\displaystyle \text{This is an angle is Quadrant 4 . . . There is another in Quadrant 3.}\)

\(\displaystyle \text{In the given interval, they are: }\;x \;=\;\begin{Bmatrix}3.5163 \\ 5.9085 \end{Bmatrix}\)

 

[whitecanvas]

Hi Soroban,

I did not realize I was solving for sinx instead of x. And I was attacking that problem for the last two days, repeating the same mistakes over and over! I got the answers from graphing it but I was so mad not being able to solve it instead of cheating with the grapher. Now I can. TQ so much!!

Aladdin,

Thanks for your time, too!

Abby

 

[stapel]

I know that : Cos(2x)= 2sin(x)cos(x) --> (1)

Actually, 2sin(x)cos(x) = sin(2x), not cos(2x). Sorry.