Question

what do you do when you have sin2x+sinx=0?

Answer 1

Are they asking if this is an identity?

Answer 2

No, it says solve the equation for solutions [0, 2pie).

Answer 3

They must want you to plug in what you have for x into that equation, because this is not an identity.

Answer 4

No, the x is just the variable they use, you really don't need it in the equation. It can be read as sin2+sin=0.

Answer 5

hmm I just finished doing something like this in calculus.

Answer 6

Lol, im in trig. And this part is on my final.

Answer 7

I see.. I just took my final! DO you mind posting a snip of this question?

Answer 8

sin2x+sinx=0
2sinxcosx+sinx=0
sin x(2cosx+1)=0
either sinx=0=sin2npi, x=2npi,where n=........,-3,-2,-1,0,1,2,3,........
or 2cosx+1=0,2cosx=-1,cosx=-1/2=-cos60=cos(180-60),cos(180+60)
x=120,240

Answer 9

we get infinite solutions of x by adding 360k from 120 and240,where k=.....,-2,-1,0,1,2,..

Answer 10

What about \[\cos2x=\sqrt{2} - \cos2x\]

Answer 11

sin2x+sinx=0 means we have to solve for x for infinite solutions which i have done

Answer 12

\[\cos 2x=\sqrt{2}-\cos 2x\]
\[\cos 2x+\cos2x=\sqrt{2}\]
\[2\cos2x=\sqrt{2}\]
\[\cos 2x=\frac{ \sqrt{2} }{ 2 }=\frac{ 1 }{\sqrt{2} }=\cos \frac{ \pi }{ 4 },\cos \frac{ -\pi }{4 }\]
\[2x=\frac{ \pi }{4 },\frac{ -\pi }{ 4 }=\frac{ \pi }{ 4 }\pm2\pi ,\frac{ -\pi }{ 4 }\pm2\pi \]
\[x=\frac{ \pi }{ 8 }\pm \pi,\frac{ -\pi }{ 8 }\pm \pi \]