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

ojchase

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Hello. I am new to this forum and need some homework help. I have been struggling with trig identities, and can't figure out how to solve sin(3x)+cos(2x)=0. I've made it this far (as x is the only variable, it is sometimes left out):

Solve sin(3x)+cos(2x)=0
(sinxcos2x+cosxsin2x)+cos2x=0
(sinxcos2x+cos(2sincos))+cos2x=0
(sincos2x+2sincos²)+cos2x=0
(sin(2cos²-1)+2sincos²)+cos2x=0
(2sincos²-sin+2sincos²)+cos2x=0
(4sincos²-sin)+cos2x=0
(sin(4cos²-1))+cos2x=0
(sin(2cos+1)(2cos-1))+cos2x=0

From there I could solve the first term alone equaling zero but it's the first term and the second term. I don't know which identity to substitute for cos(2x) as none of them appear to help me get anywhere. If anybody knows of anything I'd greatly appreciate some help.
 
Here's one of many ways.

Try getting it into sin or cos alone. I'll use sine.

\(\displaystyle \L\\sin3x+cos2x=0\)

Addition formula for sin3x:

\(\displaystyle \L\\\underbrace{sin2x}_{\text{\downarrow{2cosxsinx}}}_{\downarrow}cosx+cos2xsinx+\underbrace{1-2sin^{2}x}_{\text{cos2x}}\)

\(\displaystyle \L\\2sinxcos^{2}x+(1-2sin^{2}x)sinx+1-2sin^{2}x\)

\(\displaystyle \L\\2sinx\underbrace{(1-sin^{2}x)}_{\text{cos^2x}}+sinx-2sin^{3}x+1-2sin^{2}x\)

\(\displaystyle \L\\2sinx-2sin^{3}x+sinx-2sin^{3}x+1-2sin^{2}x\)

\(\displaystyle \L\\-4sin^{3}x+3sinx-2sin^{2}x+1\)

Let \(\displaystyle u=sinx\)

\(\displaystyle \L\\-4u^{3}-2u^{2}+3u+1=0\)

\(\displaystyle \L\\u=\frac{\sqrt{5}+1}{4}, \;\ \frac{1-\sqrt{5}}{4}, \;\ -1\)

\(\displaystyle \L\\x=sin^{-1}(\frac{\sqrt{5}+1}{4}), \;\ x=sin^{-1}(\frac{1-\sqrt{5}}{4}), \;\ x=sin^{-1}(-1)\)

Someone will come along with a more efficient way, no doubt. But, at least, this will give you identity practice.
 
Hello, ojchase!

Welcome aboard!


\(\displaystyle \sin(3x)\,+\,\cos(2x)\:=\:0\)

I have a list of multiple-angle identities now (saves a lot of time)
. . including: \(\displaystyle \,\sin(3x)\;=\:3\sin x - 4\sin^3x\,\) and \(\displaystyle \,\cos(2x)\:=\:1\,-\,2\sin^2x\)


The equation becomes: \(\displaystyle \:3\sin x\,-\,4\sin^3x \,+\,1\,-\,2\sin^2x\:=\:0\)

. . which is the cubic: \(\displaystyle \:4sin^3x \,+\,2\sin^2x\,-\,3\sin x \,-\,1\:=\:0\)

. . which factors: \(\displaystyle \:(\sin x\,+\,1)(4\sin^2x\,-\,2\sin x\,-\,1)\:=\:0\)

Now we can solve for \(\displaystyle x.\;\) (I won't solve for all solutions.)


We have: \(\displaystyle \:\sin x\,+\,1\:=\:0\;\;\Rightarrow\;\;\sin x\:=\:-1\;\;\Rightarrow\;\;\L\fbox{x\,=\,\frac{3\pi}{2}}\)


And use the Quadratic formula on: \(\displaystyle \,4\sin^2x\,-\,2\sin x \,-\,1\:=\:0\)

. . \(\displaystyle \sin x\;=\;\frac{2\,\pm\,\sqrt{20}}{8}\;=\;\frac{1\,\pm\,\sqrt{5}}{4} \;\;\Rightarrow\;\;\fbox{\L x\;=\;\sin^{-1}\left(\frac{1\,\pm\,\sqrt{5}}{4}\right)}\)

 
galactus said:
Here's one of many ways.

Try getting it into sin or cos alone. I'll use sine.

\(\displaystyle \L\\sin3x+cos2x=0\)

Addition formula for sin3x:

\(\displaystyle \L\\\underbrace{sin2x}_{\text{\downarrow{2cosxsinx}}}_{\downarrow}cosx+cos2xsinx+\underbrace{1-2sin^{2}x}_{\text{cos2x}}\)

\(\displaystyle \L\\2sinxcos^{2}x+(1-2sin^{2}x)sinx+1-2sin^{2}x\)

\(\displaystyle \L\\2sinx\underbrace{(1-sin^{2}x)}_{\text{cos^2x}}+sinx-2sin^{3}x+1-2sin^{2}x\)

\(\displaystyle \L\\2sinx-2sin^{3}x+sinx-2sin^{3}x+1-2sin^{2}x\)

\(\displaystyle \L\\-4sin^{3}x+3sinx-2sin^{2}x+1\)

Let \(\displaystyle u=sinx\)

\(\displaystyle \L\\-4u^{3}-2u^{2}+3u+1=0\)

\(\displaystyle \L\\u=\frac{\sqrt{5}+1}{4}, \;\ \frac{1-\sqrt{5}}{4}, \;\ -1\)

\(\displaystyle \L\\x=sin^{-1}(\frac{\sqrt{5}+1}{4}), \;\ x=sin^{-1}(\frac{1-\sqrt{5}}{4}), \;\ x=sin^{-1}(-1)\)

Someone will come along with a more efficient way, no doubt. But, at least, this will give you identity practice.


Everything makes sense until you substitute u in for sin...how did you factor that out (I have to show work!)?
 
I didn't factor. I just left u=sinx and solved the resulting cubic.. That means sin^3(x)=u^3. See?.
 
I done as Soroban done, multiply by -1 and get:

\(\displaystyle \L\\4u^{3}+2u^{2}-3u-1=0\)

This factors to:

\(\displaystyle \L\\(u+1)(4u^{2}-2u-1)=0\)

Now, you can see -1 is a solution. Solve the quadratic to get the other two.
 
Ah, now I understand. Thank you so much; that helped a lot. Soroban, your method works too, but I think I need to stick to the traditional identitites we've been learning. Thanks.
 
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