How Do You Solve Trigonometric Equations Involving Cosine and Sine?

[andrelutz001]

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]

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?

 

[andrelutz001]

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.

 

[Mark44]

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.