- #1
andrelutz001
- 6
- 0
Hi All,
I’m currently attempting to work ahead on my precalculs and I’m looking at trigonometric equations.
I seem to have a bit of a problem with this example (i haven’t had too many issues with the rest of the exercises),
9cos(2x)+sin(x)=9 (solve for x in the interval 0 <=x <=2pi.)
I’m thinking that I could move 9 on the other side and hence the equation will equal to 1:
cos(2x)+sin(x)=9/9
cos(2x)+sin(x)=1
I know that cos(2x)=cos2x-sin2x and than the equation should look like this:
cos2x-sin2x+sin(x)=1
Am i on the right track? What is the next step form here?
Thank you in advance.
Andrei
I’m currently attempting to work ahead on my precalculs and I’m looking at trigonometric equations.
I seem to have a bit of a problem with this example (i haven’t had too many issues with the rest of the exercises),
9cos(2x)+sin(x)=9 (solve for x in the interval 0 <=x <=2pi.)
I’m thinking that I could move 9 on the other side and hence the equation will equal to 1:
cos(2x)+sin(x)=9/9
cos(2x)+sin(x)=1
I know that cos(2x)=cos2x-sin2x and than the equation should look like this:
cos2x-sin2x+sin(x)=1
Am i on the right track? What is the next step form here?
Thank you in advance.
Andrei