Solving trigonometric equations using compound angle formula

[Tangeton]

Hello.

I find this whole topic of compound angle formula really confusing. I've been doing equations like cos(60+x) = sinx using the co function identities so far, yet this one seems to be impossible to do using cofunction identities so I need to know how to do it using compound angle formula. And yes this is definitely meant to be done using compound angle formula no double angle since in the book double angle hasnt even been mentioned at this point...

The question is to solve the equation 2sinx = 3cos(x-60) for values of x in the range 0 < x <180. I can't solve using cofunction due to the constants, but if anyone knows how to solve using cofunction identities with the constants this kind of answer is also welcomed.

I tried using compound angle formula after using cofunction identities where 2sinx = 2cos(90-x)

I tried to bring everything to the power of 2 but this gave me 11tan^2x = -9 so that is impossible.
I tried to take away 3sqrt3tan from 4tan which equaled 3 but this gave me the wrong answer.

The answer is 111.7.

Ive used x and feta interchangeably as don't know how to place feta symbol on keyboard.

Thanks for help.

EDIT 2: Corrected every typo eveywhere

 

[PeroK]

Hello.

I find this whole topic of compound angle formula really confusing. I've been doing equations like cos(60+x) = sinx using the co function identities so far, yet this one seems to be impossible to do using cofunction identities so I need to know how to do it using compound angle formula. And yes this is definitely meant to be done using compound angle formula no double angle since in the book double angle hasnt even been mentioned at this point...

The question is to solve the equation 2sinx = 3cos(x-60) for values of x in the range 0 < x <180. I can't solve using cofunction due to the constants, but if anyone knows how to solve using cofunction identities with the constants this kind of answer is also welcomed.

I tried using compound angle formula after using cofunction identities where 2sinx = 2cos(90-x)

I tried to take away 3sqrt3tan from 4tan which equaled 3 but this gave me the wrong answer.

The answer is 111.7.

Why did this give you the wrong answer?

 

[DEvens]

Nearly there. In your last equation just solve for $tan \theta$.

 

[ehild]

Hello.

I find this whole topic of compound angle formula really confusing. I've been doing equations like cos(60+x) = sinx using the co function identities so far, yet this one seems to be impossible to do using cofunction identities so I need to know how to do it using compound angle formula. And yes this is definitely meant to be done using compound angle formula no double angle since in the book double angle hasnt even been mentioned at this point...

The question is to solve the equation 2sinx = 3cos(x-60) for values of x in the range 0 < x <180. I can't solve using cofunction due to the constants, but if anyone knows how to solve using cofunction identities with the constants this kind of answer is also welcomed.

I tried using compound angle formula after using cofunction identities where 2sinx = 2cos(90-x)

I tried to bring everything to the power of 2

Why? You know what √3 is.
By the way, you lost a "3". After multiplication by 2 , the right hand side is 3cos(θ)+3√3 sin(θ), so you get the equation 4tan(θ)=3+3√3tan(θ) for tan(θ).
Bring the tan(θ) terms on one side, solve for tan(θ).

 

[Tangeton]

4tan(θ)=3+3√3tan(θ)
3+3√3tan(θ)- 4tan(θ) = 0
3 + (3√3-4)tan(θ) = 0
(3√3-4)Tan(θ) = -3
Tan(θ) = -3/(3√3-4)
θ = Tan^-1(-3/(3√3-4) ) = -68.3 (3sf) but the answer is 111.7...

And upon plugging in x = -68.3 and x = 111.7, it turns out that they both do make the equation 2sinx = 3cos(x-60) equal on both sides, yet x = -68.3 gives equal negative and x = 111.7 gives equal positives. 180 - 68.3 = 111.7. So I found the reason for my result. Would you guys say both answers would be accepted...?

EDIT: I actually got the same answer before making this post but book said it was wrong so I got really confused.

 

[PeroK]

4tan(θ)=3+3√3tan(θ)
3+3√3tan(θ)- 4tan(θ) = 0
3 + (3√3-4)tan(θ) = 0
(3√3-4)Tan(θ) = -3
Tan(θ) = -3/(3√3-4)
θ = Tan^-1(-3/(3√3-4) ) = -68.3 (3sf) but the answer is 111.7...

And upon plugging in x = -68.3 and x = 111.7, it turns out that they both do make the equation 2sinx = 3cos(x-60) equal on both sides, yet x = -68.3 gives equal negative and x = 111.7 gives equal positives. 180 - 68.3 = 111.7. So I found the reason for my result. Would you guys say both answers would be accepted...?

Not if $0 < x < 180$ as was stated in your original post.

 

[Tangeton]

Not if $0 < x < 180$ as was stated in your original post.

Ahh okay I forgot about that bit.. I guess I would have to remember to use the CAST diagram for my answer. Thanks for all help, everyone.

 

[ehild]

4tan(θ)=3+3√3tan(θ)
3+3√3tan(θ)- 4tan(θ) = 0
3 + (3√3-4)tan(θ) = 0
(3√3-4)Tan(θ) = -3
Tan(θ) = -3/(3√3-4)
θ = Tan^-1(-3/(3√3-4) ) = -68.3 (3sf) but the answer is 111.7...

You know that the tangent function is periodic with period of 180°. And you need to find solution in the range 0<θ<180°. Adding 180° to your result is also solution, just in the desired range.