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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