How to solve trignometric equations ?

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The discussion focuses on solving trigonometric equations, specifically two examples provided by the original poster. The first equation, 2sin(P) + 1.5cos(P) = 10, is noted to have no real solutions since the maximum value of the left side can only reach 3.5. The second equation, cos(P) + 50 = tan(P), also presents challenges, leading to a complex polynomial form that complicates finding solutions. Participants suggest using the method of expressing Acosθ + Bsinθ in the form Csin(θ + φ) but emphasize the need for clarity on how to derive values for C and φ. Overall, the thread highlights the difficulties in solving these equations and the importance of understanding trigonometric identities and numerical techniques.
bongas
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Hello maths masters...

I don't know how to solve trignometric equations, let me give you people a few examples and then any of you can guide me through it by :
1)Giving a link to a good site
2)by solving the examples provided (and also explaining them)
3)or simply by stating(an easy to understand) formula.

i don't have a mathemetics keyboard or software so consider " P " as a variable whose value is to be found.

The example 1:
2sin(P) + 1.5cos(P) = 10
The example 2:
cos(P) + 50 = tan(P)


Please solve both of the examples and also show me the working so that i can clearly understand how its done.
 
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You should be familiar with the formula sin^2x+cos^2x=1

Well, re-arranging this formula in some ways can give you things like

cos^2x=1-sin^2x

and

sinx=\pm \sqrt{1-cos^2x}

Also remember that tanx=\frac{sinx}{cosx}
 
hi bongas! :smile:

(have a theta: θ :wink:)
bongas said:
2sin(P) + 1.5cos(P) = 10

hint: Acosθ + Bsinθ can be written in the form Csin(θ + φ) :wink:
cos(P) + 50 = tan(P)

i don't know a simple solution to this, other than using Mentallic's :smile: hint
 
bongas said:
the answer u gave was enough for solving example no.1 but i see no way of solving example no. 2


by using what the general rules of trignometric ratios you mentioned i could only go as far as :

cos^4(P) + 51cos^2(P) - 1 = 0

Now, how do we solve this ?


Note that cos^4 means cosine raise to the power 2 or in other words cosine squared.

I didn't get that result for either question 1 or 2. Are you expected to use numerical techniques in this class? Because they don't seem to be giving nice answers.
 
tiny-tim said:
hint: Acosθ + Bsinθ can be written in the form Csin(θ + φ) :wink:

Hopefully the OP can notice that no real solutions exist by using common sense :smile:
 
Mentallic said:
Hopefully the OP can notice that no real solutions exist by using common sense :smile:

I assumed bongas just put in numbers at random, as an example. :wink:
 
oh yes these were just random numbers, it was not a proven question...but i have already told the bedrock of my question ie can anyone explain me how these type of equations are solved.
@tiny tim, i have seen this methode being used very often but have never really understood how to find the value of (phi) and (C) ? It would be really helpful if you show it to me (the working ) of any proven example.
 
Mentallic said:
I didn't get that result for either question 1 or 2. Are you expected to use numerical techniques in this class? Because they don't seem to be giving nice answers.

What numerical techniques are you talking about ?
Can you please specify and also it would be really helpfull if you show me the difference between the two techniques .
 
hi bongas! :wink:
bongas said:
@tiny tim, i have seen this methode being used very often but have never really understood how to find the value of (phi) and (C) ?

(what happened to that φ i gave you? :redface:)

hint: expand Csin(θ + φ) …

what do you get? :smile:
 
  • #10
By observation it can be seen that there are no real solutions to the first equation.

Sin and Cos both have a max of 1, so the biggest the left side can be is 3.5, it can never, in the real numbers equal 10.
 
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