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# using acosx+bsinx=c watch

1. The question is to solve the equation 2cos2x +sqrt5 sin2x =3 between -π and π
I used the equation Rsin(2x+α) I found R to be 3 and α to be 0.841
I'm having trouble solving this equation, not sure whether it is beacuse of the R and α values.

2. Your alpha is a bit out.
3. (Original post by Mr M)
Your alpha is a bit out.
how will I work out alpha? I've tried working it out but not what I've done wrong :/
4. (Original post by Lavanyaa)
how will I work out alpha? I've tried working it out but not what I've done wrong :/
Expand ?
5. (Original post by Mr M)
Expand ?
As the equation is equal to 3 don't you need to take the 3 to the other side and divide it by the other 3 ?
6. (Original post by Lavanyaa)
As the equation is equal to 3 don't you need to take the 3 to the other side and divide it by the other 3 ?
You need to find alpha first.

Do you actually understand how to express a number in harmonic form?
7. (Original post by Mr M)
You need to find alpha first.

Do you actually understand how to express a number in harmonic form?
I wasn't taught this way, is it easier?
8. (Original post by Lavanyaa)
I wasn't taught this way, is it easier?
Easier than what? Show your working.

The only alternative is making a half tangent substitution. That is a little quicker but I'm fairly sure you are supposed to be using the line of attack you proposed in your question.
9. (Original post by Mr M)
Easier than what? Show your working.

The only alternative is making a half tangent substitution. That is a little quicker but I'm fairly sure you are supposed to be using the line of attack you proposed in your question.
I found α to be 0.7297...
so one of the values will be 0.4205
10. (Original post by Lavanyaa)
I found α to be 0.7297...
so one of the values will be 0.4205
yes

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