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Need help with trigonometric equation

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
 

danago

Gold Member
1,122
4
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
Woa be careful what you do there; If you are going to divide by 9, make sure you divide EVERYTHING by 9, so it becomes:

cos(2x) + (1/9)sin(x)=1

Anyway, other than that, i would probably proceed in the same way as you did. With these types of questions, it is often easiest to reduce the equation to a form such that only one type of trig function is present. Can you see how a common identity can be used to further reduce the equation down to one with only sine's? What common type of equation does this then resemble?
 
Thank you for replaying danago.

That definitely helps. So after dividing everything by 9 I’m getting:
cos(2x)+1/9sin(x)=1
I can use the double angle formula to reduce the equation to one type of trig function, hence:
1-2sin^2(x)+1/9sin(x)=1
-sin^2(x)+1/9sin(x)=0
And i now have a quadratic equation type 2x^2+(1/9)x
I can use the quadratic formula and I’m nearly done.

Many thanks.
 
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4,301
Two comments:
When you write cos(2x)+1/9sin(x)=1, some people might (incorrectly) take the sine term to be 1/(9sin(x)). You can write this more clearly as 1/9 * sin(x).

You can use the quadratic formula to solve -sin^2(x)+1/9sin(x)=0, but it's quicker and simpler just to factor sin(x) from each term to get sin(x)(-sin(x) + 1/9) = 0, and then set each factor to 0 to solve for sin(x) and then x.
 

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